🧠:[<think>Alright, let me try to work through this problem step by step. It's about an ellipse and a circle tangent to a line, then some geometry with points on the ellipse. Let's take it part by part.First, problem (1): Find the equation of the ellipse C. The ellipse is given by x²/a² + y²/b² = 1, with a > b > 0, eccentricity 1/2. There's also a circle centered at the origin with radius equal to the minor axis of the ellipse (which is b, right? Because for an ellipse, the minor axis is the shorter one, so length 2b, hence radius b). This circle is tangent to the line x - y + √6 = 0. So, I need to find a and b such that the ellipse has eccentricity 1/2 and the circle with radius b is tangent to that line.Starting with the circle. The circle is x² + y² = b². The line is x - y + √6 = 0. The distance from the center of the circle (which is the origin) to the line should be equal to the radius for them to be tangent. The distance from a point (x₀, y₀) to the line Ax + By + C = 0 is |Ax₀ + By₀ + C| / √(A² + B²). Here, the line is x - y + √6 = 0, so A = 1, B = -1, C = √6. The center is (0,0), so distance is |0 + 0 + √6| / √(1 + 1) = √6 / √2 = √(6/2) = √3. Therefore, the radius of the circle is √3. But the radius is equal to the minor axis of the ellipse, which is b. So b = √3.Now, since the ellipse has eccentricity e = 1/2. The formula for eccentricity of an ellipse is e = c/a, where c is the distance from the center to a focus, and c = √(a² - b²). So, e = √(a² - b²)/a = 1/2. Let's plug in b = √3.√(a² - (√3)²)/a = 1/2√(a² - 3)/a = 1/2Square both sides:(a² - 3)/a² = 1/4Multiply both sides by a²:a² - 3 = a²/4Subtract a²/4 from both sides:(3/4)a² - 3 = 03a²/4 = 3Multiply both sides by 4/3:a² = 4So, a = 2 (since a > 0). Therefore, the equation of the ellipse is x²/4 + y²/3 = 1.Let me check that again. The circle radius is b = √3, and the distance from origin to the line x - y + √6 = 0 is √3, which matches. Then, eccentricity e = c/a = √(a² - b²)/a. Plugging a² = 4 and b² = 3, we get √(4 - 3)/2 = √1/2 = 1/2. Perfect, that matches the given eccentricity. So part (1) is done. The ellipse equation is x²/4 + y²/3 = 1.Now part (2): A line passing through the fixed point Q(1, 0) with slope k intersects the ellipse C at points M and N. If the dot product of vectors OM and ON is -2, find the value of the real number k and the area of triangle MON.Okay, so we need to find k such that when we draw a line through (1,0) with slope k, intersecting the ellipse at M and N, then the dot product OM • ON = -2, and then find the area of triangle MON.First, let's recall that the dot product of vectors OM and ON is equal to (x1x2 + y1y2) where M is (x1, y1) and N is (x2, y2). So, we need that x1x2 + y1y2 = -2.So, the strategy is: write the equation of the line with slope k passing through (1,0). Find its intersection points with the ellipse. Then compute x1x2 + y1y2 and set it equal to -2. Then solve for k. Then use that to find the area.Let me start by writing the equation of the line. It passes through (1,0) with slope k. So, the equation is y = k(x - 1).We can substitute this into the ellipse equation x²/4 + y²/3 = 1 to find the intersection points.Substituting y = k(x - 1) into the ellipse:x²/4 + [k²(x - 1)²]/3 = 1Multiply both sides by 12 to eliminate denominators:3x² + 4k²(x - 1)² = 12Expand the (x - 1)² term:3x² + 4k²(x² - 2x + 1) = 12Expand further:3x² + 4k²x² - 8k²x + 4k² = 12Combine like terms:(3 + 4k²)x² - 8k²x + (4k² - 12) = 0So, this is a quadratic in x. Let's denote this as:A x² + B x + C = 0, whereA = 3 + 4k²B = -8k²C = 4k² - 12Let the roots be x1 and x2. Then, by Vieta's formula:x1 + x2 = -B/A = 8k² / (3 + 4k²)x1 x2 = C/A = (4k² - 12)/(3 + 4k²)Similarly, since y = k(x - 1), then y1 = k(x1 - 1), y2 = k(x2 - 1)Therefore, y1 y2 = k² (x1 - 1)(x2 - 1) = k² [x1x2 - (x1 + x2) + 1]So, x1x2 + y1y2 = x1x2 + k² [x1x2 - (x1 + x2) + 1]Let me compute this expression step by step.First, compute x1x2: (4k² - 12)/(3 + 4k²)Compute x1 + x2: 8k²/(3 + 4k²)Then, x1x2 - (x1 + x2) + 1 = [(4k² - 12)/(3 + 4k²)] - [8k²/(3 + 4k²)] + 1Combine terms:[ (4k² - 12) - 8k² + (3 + 4k²) ] / (3 + 4k²)Compute numerator:4k² -12 -8k² +3 +4k² = (4k² -8k² +4k²) + (-12 +3) = 0k² -9 = -9Therefore, x1x2 - (x1 + x2) +1 = -9/(3 +4k²)Therefore, y1y2 = k²*(-9)/(3 +4k²) = -9k²/(3 +4k²)Therefore, x1x2 + y1y2 = [ (4k² -12)/(3 +4k²) ] + [ -9k²/(3 +4k²) ]Combine numerator:(4k² -12 -9k²)/(3 +4k²) = (-5k² -12)/(3 +4k²)But according to the problem, this sum is equal to -2. Therefore:(-5k² -12)/(3 +4k²) = -2Multiply both sides by (3 +4k²):-5k² -12 = -2(3 +4k²)Expand RHS:-5k² -12 = -6 -8k²Bring all terms to left side:-5k² -12 +6 +8k² = 0Simplify:3k² -6 =03k² =6k²=2Thus, k=√2 or k=-√2. So k is ±√2.Wait, that's the answer? Let me check again.Starting from x1x2 + y1y2 = -2Computed x1x2 + y1y2 = (-5k² -12)/(3 +4k²) = -2So, solving:(-5k² -12) = -2*(3 +4k²)-5k² -12 = -6 -8k²Adding 8k² +6 to both sides:3k² -6 =03k² =6 => k²=2 => k=±√2. Yes, that's correct.Therefore, the values of k are √2 and -√2.Next, find the area of triangle MON.Triangle MON has vertices at O(0,0), M(x1,y1), N(x2,y2). The area can be computed using the determinant formula:Area = (1/2)|x1 y2 - x2 y1|Alternatively, since the area is (1/2)|OM × ON|, the magnitude of the cross product divided by 2.But the cross product in 2D is scalar x1y2 - x2y1. So, Area = (1/2)|x1y2 - x2y1|Alternatively, if we can express this in terms of the coordinates.Alternatively, maybe we can find the length of MN and the height from O, but perhaps the determinant is easier.But to compute x1y2 - x2y1, we can note that since y1 = k(x1 -1) and y2 =k(x2 -1), so:x1y2 -x2y1 = x1*k(x2 -1) -x2*k(x1 -1) = k[ x1x2 -x1 -x1x2 +x2 ] = k[ -x1 +x2 ]= k(x2 -x1)Therefore, Area = (1/2)|k(x2 -x1)|But (x2 -x1) is the difference of roots. From quadratic equation, we know that (x2 -x1) = sqrt[(x1 +x2)^2 -4x1x2]. But actually, the difference x2 -x1 can be positive or negative, but since we take absolute value, the area is (1/2)|k| * |x2 -x1|So, let's compute |x2 -x1|. From quadratic equation, the discriminant D = B² -4AC.Earlier, quadratic was A x² + Bx + C =0, with A=3 +4k², B=-8k², C=4k² -12.Discriminant D = (-8k²)^2 -4*(3 +4k²)*(4k² -12)Compute D:64k⁴ -4*(3 +4k²)*(4k² -12)First compute the product (3 +4k²)(4k² -12):= 3*4k² +3*(-12) +4k²*4k² +4k²*(-12)=12k² -36 +16k⁴ -48k²=16k⁴ -36k² -36Multiply by -4:-4*(16k⁴ -36k² -36) = -64k⁴ +144k² +144Therefore, discriminant D=64k⁴ -64k⁴ +144k² +144 = 144k² +144 = 144(k² +1)Therefore, sqrt(D) = 12*sqrt(k² +1)Therefore, |x2 -x1| = sqrt(D)/|A| = 12√(k² +1)/ (3 +4k²)Since we have k²=2, then k=±√2, so k²=2.Plugging k²=2:|x2 -x1| =12√(2 +1)/(3 +4*2) =12√3/(3 +8)=12√3/11Therefore, |x2 -x1|=12√3/11Then, |k|=√2Therefore, Area=(1/2)*√2*(12√3/11)= (1/2)*(12√6)/11= (6√6)/11Hence, the area is 6√6/11.Let me check this again.First, discriminant D=144(k² +1). So sqrt(D)=12sqrt(k² +1). Then |x2 -x1|=sqrt(D)/A where A=3 +4k². So 12sqrt(k² +1)/(3 +4k²). Then, for k²=2, this is 12sqrt(3)/(3 +8)=12sqrt(3)/11.Then, Area=(1/2)*|k|*|x2 -x1|. So |k|=sqrt(2), so (1/2)*sqrt(2)*(12sqrt(3)/11)= (12/11)*(sqrt(6)/2)=6sqrt(6)/11. Yes, that's correct.So the area is 6√6/11.Therefore, the answers are:(1) The ellipse equation is x²/4 + y²/3 = 1(2) k = ±√2, and the area is 6√6/11.I need to confirm all steps again.First part (1) seems correct. The circle with radius b (minor axis) is tangent to the line, distance from origin is √3, so b=√3. Eccentricity e=1/2=√(a² -b²)/a. Solving gives a=2. So equation x²/4 + y²/3=1.Second part (2): The line through Q(1,0) with slope k: y=k(x-1). Substituted into ellipse, found quadratic equation. Then using Vieta's, found x1x2 and y1y2 in terms of k, set x1x2 + y1y2=-2, solved for k, got k=±√2. Then computed area using determinant, expressed in terms of |x2 -x1|, found |x2 -x1|=12√3/11 when k²=2, then Area=(1/2)*|k|*|x2 -x1|=6√6/11. All steps seem correct.Final Answer(1) The equation of the ellipse is boxed{dfrac{x^2}{4} + dfrac{y^2}{3} = 1}.(2) The value of ( k ) is boxed{pm sqrt{2}} and the area of ( triangle MON ) is boxed{dfrac{6sqrt{6}}{11}}.</think>Given an ellipse ( C: frac{x^2}{a^2} + frac{y^2}{b^2} = 1 ) with eccentricity ( frac{1}{2} ), and a circle centered at the origin with radius equal to the minor axis of the ellipse (radius ( b )) that is tangent to the line ( x - y + sqrt{6} = 0 ):1. Finding the equation of the ellipse ( C ): - The distance from the origin to the line ( x - y + sqrt{6} = 0 ) is calculated as ( frac{sqrt{6}}{sqrt{2}} = sqrt{3} ), which is the radius of the circle. Thus, ( b = sqrt{3} ). - The eccentricity ( e = frac{1}{2} ) gives ( frac{sqrt{a^2 - b^2}}{a} = frac{1}{2} ). Solving this with ( b = sqrt{3} ): [ frac{sqrt{a^2 - 3}}{a} = frac{1}{2} implies sqrt{a^2 - 3} = frac{a}{2} implies a^2 - 3 = frac{a^2}{4} implies frac{3a^2}{4} = 3 implies a^2 = 4 implies a = 2 ] - Therefore, the equation of the ellipse is ( frac{x^2}{4} + frac{y^2}{3} = 1 ).2. Finding the value of ( k ) and the area of ( triangle MON ): - The line through ( Q(1, 0) ) with slope ( k ) is ( y = k(x - 1) ). Substituting into the ellipse equation: [ frac{x^2}{4} + frac{[k(x - 1)]^2}{3} = 1 implies 3x^2 + 4k^2(x^2 - 2x + 1) = 12 implies (3 + 4k^2)x^2 - 8k^2x + (4k^2 - 12) = 0 ] - Using Vieta's formulas, the product ( x_1x_2 = frac{4k^2 - 12}{3 + 4k^2} ) and the sum ( x_1 + x_2 = frac{8k^2}{3 + 4k^2} ). The product ( y_1y_2 ) is calculated as: [ y_1y_2 = k^2 left( x_1x_2 - (x_1 + x_2) + 1 right) = k^2 left( frac{-9}{3 + 4k^2} right) ] - Setting ( x_1x_2 + y_1y_2 = -2 ): [ frac{-5k^2 - 12}{3 + 4k^2} = -2 implies k^2 = 2 implies k = pm sqrt{2} ] - The area of ( triangle MON ) is calculated using the determinant formula. The discriminant of the quadratic equation gives ( |x_2 - x_1| = frac{12sqrt{3}}{11} ). Thus, the area is: [ text{Area} = frac{1}{2} |k| |x_2 - x_1| = frac{1}{2} sqrt{2} cdot frac{12sqrt{3}}{11} = frac{6sqrt{6}}{11} ] Final Answer(1) The equation of the ellipse is boxed{dfrac{x^2}{4} + dfrac{y^2}{3} = 1}.(2) The value of ( k ) is boxed{pm sqrt{2}} and the area of ( triangle MON ) is boxed{dfrac{6sqrt{6}}{11}}.,<think>Okay, let's tackle this problem step by step. It's about an ellipse and a circle tangent to a line, then some geometry involving points on the ellipse. Let me start with part (1), finding the equation of the ellipse C.First, the given ellipse has the standard equation ( frac{x^2}{a^2} + frac{y^2}{b^2} = 1 ), where ( a > b > 0 ). The eccentricity is given as ( frac{1}{2} ). I remember that the eccentricity e of an ellipse is related to a and b by ( e = frac{c}{a} ), where c is the distance from the center to each focus, and ( c^2 = a^2 - b^2 ). So since e = 1/2, we have ( c = frac{a}{2} ). Then substituting into ( c^2 = a^2 - b^2 ):( left(frac{a}{2}right)^2 = a^2 - b^2 )Simplifying:( frac{a^2}{4} = a^2 - b^2 )Multiply both sides by 4 to eliminate the denominator:( a^2 = 4a^2 - 4b^2 )Bring all terms to one side:( -3a^2 + 4b^2 = 0 )Wait, that's ( 4b^2 = 3a^2 ), so ( b^2 = frac{3}{4}a^2 ). Hmm, so that relates b and a. But we need another equation to find the specific values of a and b. The next part of the problem mentions a circle centered at the origin with radius equal to the minor axis of the ellipse, which is b, and this circle is tangent to the line ( x - y + sqrt{6} = 0 ).So, the circle has equation ( x^2 + y^2 = b^2 ), and it's tangent to the line ( x - y + sqrt{6} = 0 ). The condition for tangency between a line and a circle is that the distance from the center of the circle to the line equals the radius of the circle. Since the circle is centered at the origin, the distance from (0,0) to the line ( x - y + sqrt{6} = 0 ) should be equal to b.The formula for the distance from a point (x₀,y₀) to the line Ax + By + C = 0 is ( frac{|Ax₀ + By₀ + C|}{sqrt{A^2 + B^2}} ). Here, the line is ( 1x - 1y + sqrt{6} = 0 ), so A = 1, B = -1, C = √6. The center is (0,0), so substituting:Distance = ( frac{|0 + 0 + sqrt{6}|}{sqrt{1 + 1}} = frac{sqrt{6}}{sqrt{2}} = sqrt{3} )So the radius b is equal to √3. Therefore, b = √3. Then from earlier, we have ( b^2 = 3 = frac{3}{4}a^2 ), so solving for a²:( 3 = frac{3}{4}a^2 )Multiply both sides by 4/3:( a^2 = 4 )Therefore, a = 2 (since a > 0). So the equation of the ellipse is ( frac{x^2}{4} + frac{y^2}{3} = 1 ). Let me just check if that's correct.Eccentricity e = c/a, where c = √(a² - b²) = √(4 - 3) = √1 = 1. So e = 1/2, which matches the given. The circle with radius b = √3 is tangent to the line x - y + √6 = 0. The distance from the origin to the line is √3, which equals the radius. So that's correct. So part (1) answer is ( frac{x^2}{4} + frac{y^2}{3} = 1 ).Now part (2). A line passes through Q(1, 0) with slope k, intersects the ellipse C at points M and N. Given that the dot product of vectors OM and ON is -2, find k and the area of triangle MON.First, let's parametrize the line. The line passes through Q(1, 0) with slope k, so its equation is y = k(x - 1). We need to find where this line intersects the ellipse. Substitute y = k(x - 1) into the ellipse equation:( frac{x^2}{4} + frac{[k(x - 1)]^2}{3} = 1 )Expand that:( frac{x^2}{4} + frac{k^2(x^2 - 2x + 1)}{3} = 1 )Multiply through by 12 to eliminate denominators:3x² + 4k²(x² - 2x + 1) = 12Expand:3x² + 4k²x² - 8k²x + 4k² = 12Combine like terms:(3 + 4k²)x² - 8k²x + (4k² - 12) = 0This is a quadratic equation in x. Let me write it as:(4k² + 3)x² - 8k²x + (4k² - 12) = 0Let’s denote coefficients:A = 4k² + 3B = -8k²C = 4k² - 12Let the roots be x₁ and x₂, corresponding to points M and N. Then, from quadratic equation:x₁ + x₂ = -B/A = (8k²)/(4k² + 3)x₁x₂ = C/A = (4k² - 12)/(4k² + 3)Similarly, the y-coordinates of M and N are y₁ = k(x₁ - 1) and y₂ = k(x₂ - 1)Now, the vectors OM and ON are (x₁, y₁) and (x₂, y₂). Their dot product is:OM • ON = x₁x₂ + y₁y₂ = x₁x₂ + [k(x₁ - 1)][k(x₂ - 1)]Given that this equals -2. Let's compute that:First, compute y₁y₂:k²(x₁ - 1)(x₂ - 1) = k²[x₁x₂ - x₁ - x₂ + 1]Therefore, OM • ON = x₁x₂ + k²(x₁x₂ - x₁ - x₂ + 1)So, substitute x₁ + x₂ and x₁x₂ from above:OM • ON = x₁x₂ + k²(x₁x₂ - (x₁ + x₂) + 1)Let me compute each term:Let’s denote S = x₁ + x₂ = (8k²)/(4k² + 3)P = x₁x₂ = (4k² - 12)/(4k² + 3)So, OM • ON = P + k²(P - S + 1)Substitute S and P:= P + k²(P - S + 1)= (4k² - 12)/(4k² + 3) + k²[ (4k² - 12)/(4k² + 3) - (8k²)/(4k² + 3) + 1 ]Let me compute the expression inside the brackets:(4k² - 12 - 8k²)/(4k² + 3) + 1 = (-4k² -12)/(4k² + 3) + 1= [ -4k² -12 + 4k² + 3 ] / (4k² + 3 )= (-9)/(4k² + 3)Therefore, OM • ON becomes:(4k² -12)/(4k² +3) + k²*(-9)/(4k² +3 )= [4k² -12 -9k²]/(4k² +3 )= (-5k² -12)/(4k² +3 )Set this equal to -2:(-5k² -12)/(4k² +3 ) = -2Multiply both sides by (4k² +3):-5k² -12 = -2(4k² +3 )Simplify RHS:-5k² -12 = -8k² -6Bring all terms to left side:-5k² -12 +8k² +6 =03k² -6=03k² =6k²=2Therefore, k=√2 or k=-√2. Since the problem says "real number k", both are real. So k=±√2.Now, need to find the area of triangle MON. The area of triangle formed by points O(0,0), M(x₁,y₁), N(x₂,y₂) is (1/2)|x₁y₂ -x₂y₁|.We can compute this determinant. Let’s compute x₁y₂ - x₂y₁.Since y₁ =k(x₁ -1), y₂=k(x₂ -1)So x₁y₂ -x₂y₁ =x₁[k(x₂ -1)] -x₂[k(x₁ -1)] =k[x₁x₂ -x₁ -x₁x₂ +x₂] =k[ -x₁ +x₂ ] =k(x₂ -x₁)Therefore, the area is (1/2)|k(x₂ -x₁)| = (1/2)|k| |x₂ -x₁|To find |x₂ -x₁|, we can use the formula for the difference of roots in quadratic equation:|x₂ -x₁| = √[(x₁ +x₂)^2 -4x₁x₂] = √[S² -4P]We have S =8k²/(4k² +3), P=(4k² -12)/(4k² +3)Compute S² -4P:= [64k^4/(4k² +3)^2] -4*(4k² -12)/(4k² +3)= [64k^4 -4*(4k² -12)(4k² +3)]/(4k² +3)^2Let me compute the numerator:64k^4 -4*(4k² -12)(4k² +3)First, expand (4k² -12)(4k² +3):=16k^4 +12k² -48k² -36=16k^4 -36k² -36Multiply by 4:=64k^4 -144k² -144Therefore, numerator:64k^4 - (64k^4 -144k² -144) =64k^4 -64k^4 +144k² +144 =144k² +144=144(k² +1)Therefore, S² -4P =144(k² +1)/(4k² +3)^2Therefore, |x₂ -x₁|=√[144(k² +1)/(4k² +3)^2] =12√(k² +1)/(4k² +3)Therefore, area is (1/2)|k| *12√(k² +1)/(4k² +3) =6|k|√(k² +1)/(4k² +3)Since k=±√2, |k|=√2, k²=2Substitute:6*√2*√(2 +1)/(4*2 +3) =6√2*√3/(8 +3)=6√6/11So area is 6√6/11.Wait, let me check the calculation step by step.First, with k=±√2, |k|=√2, k²=2.Compute numerator:6|k|√(k² +1) =6√2 *√(2 +1)=6√2*√3=6√6Denominator:4k² +3=4*2 +3=8 +3=11Thus, area=6√6/11.Therefore, the area of triangle MON is 6√6/11.So summarizing part (2): k=±√2 and area=6√6/11.Let me check if all steps are correct.Starting from the line equation, substituting into ellipse, getting quadratic in x, finding expressions for S and P, then computing OM•ON, setting it equal to -2, solving for k. Then computing the area using determinant. Looks good. The key steps are correct. So I think this is right.Final Answer(1) The equation of the ellipse ( C ) is boxed{dfrac{x^2}{4} + dfrac{y^2}{3} = 1}.(2) The value of ( k ) is boxed{pm sqrt{2}} and the area of ( triangle MON ) is boxed{dfrac{6sqrt{6}}{11}}.</think>Given an ellipse ( C: frac{x^2}{a^2} + frac{y^2}{b^2} = 1 ) with eccentricity ( frac{1}{2} ), and a circle centered at the origin with radius equal to the minor axis of the ellipse is tangent to the line ( x - y + sqrt{6} = 0 ):1. Finding the equation of the ellipse ( C ): - The eccentricity ( e = frac{1}{2} ) gives ( c = frac{a}{2} ). - Using ( c^2 = a^2 - b^2 ), we get ( left(frac{a}{2}right)^2 = a^2 - b^2 ), leading to ( a^2 = 4b^2/3 ). - The circle with radius ( b ) is tangent to the line ( x - y + sqrt{6} = 0 ). The distance from the origin to the line is ( sqrt{3} ), so ( b = sqrt{3} ). - Substituting ( b = sqrt{3} ) into ( a^2 = 4b^2/3 ), we get ( a^2 = 4 ). Thus, ( a = 2 ). - The equation of the ellipse is ( frac{x^2}{4} + frac{y^2}{3} = 1 ).2. Finding the value of ( k ) and the area of ( triangle MON ): - The line through ( Q(1, 0) ) with slope ( k ) is ( y = k(x - 1) ). - Substituting this into the ellipse equation gives a quadratic equation in ( x ): ( (4k^2 + 3)x^2 - 8k^2x + (4k^2 - 12) = 0 ). - Using Vieta's formulas, the roots ( x_1 ) and ( x_2 ) give ( x_1 + x_2 = frac{8k^2}{4k^2 + 3} ) and ( x_1x_2 = frac{4k^2 - 12}{4k^2 + 3} ). - The dot product ( overrightarrow{OM} cdot overrightarrow{ON} = -2 ) leads to solving ( frac{-5k^2 - 12}{4k^2 + 3} = -2 ), resulting in ( k = pm sqrt{2} ). - The area of ( triangle MON ) is calculated using the determinant formula, resulting in ( frac{6sqrt{6}}{11} ).Final Answers:1. The equation of the ellipse ( C ) is boxed{dfrac{x^2}{4} + dfrac{y^2}{3} = 1}.2. The value of ( k ) is boxed{pm sqrt{2}} and the area of ( triangle MON ) is boxed{dfrac{6sqrt{6}}{11}}.]

🔑: Step-by-Step Solution# Part 1: Finding the Equation of the Ellipse CGiven the eccentricity of the ellipse C is e=frac{c}{a}=frac{1}{2}, we can express this relationship as follows:1. {e}^{2}=frac{{c}^{2}}{{a}^{2}}=frac{{a}^{2}-{b}^{2}}{{a}^{2}}=frac{1}{4}This implies:2. {a}^{2}=frac{4}{3}{b}^{2}Given that the circle odot E with center at the origin and radius equal to the minor axis of the ellipse is tangent to the line x-y+sqrt{6}=0, we find the radius b as follows:3. b=frac{sqrt{6}}{sqrt{1+1}}=sqrt{3}Solving equations from steps 2 and 3, we get:4. a^{2}=4 and b^{2}=3Thus, the equation of the ellipse C is:boxed{frac{x^{2}}{4}+frac{y^{2}}{3}=1}# Part 2: Finding the Value of k and the Area of triangle MONLet the equation of the line MN be y=k(x-1), with M(x_{1}, y_{1}) and N(x_{2}, y_{2}). Solving the system:1. left{begin{array}{c}frac{{x}^{2}}{4}+frac{{y}^{2}}{3}=1 y=k(x-1)end{array}right.Eliminating y and simplifying, we get:2. (3+4k^{2})x^{2}-8k^{2}x+4k^{2}-12=0Applying Vieta's formulas, we have:3. left{begin{array}{c}{x}_{1}+{x}_{2}=frac{8{k}^{2}}{3+4{k}^{2}}{x}_{1}•{x}_{2}=frac{4{k}^{2}-12}{3+4{k}^{2}}end{array}right.Then, {y}_{1}{y}_{2}=k({x}_{1}-1)•k({x}_{2}-1)=frac{-9{k}^{2}}{3+4{k}^{2}}.Given overrightarrow{OM}•overrightarrow{ON}={x}_{1}{x}_{2}+{y}_{1}{y}_{2}=frac{-5{k}^{2}-12}{3+4{k}^{2}}=-2, we solve for k^{2}:4. k^{2}=2Solving for k, we find:5. k=pmsqrt{2}Substituting k back into the system, we get:6. left{begin{array}{c}{x}_{1}+{x}_{2}=frac{16}{11}{x}_{1}•{x}_{2}=frac{-4}{11}end{array}right.Thus, |MN|=sqrt{{({x}_{1}-{x}_{2})}^{2}+{({y}_{1}-{y}_{2})}^{2}}=sqrt{1+{k}^{2}}•sqrt{{({x}_{1}+{x}_{2})}^{2}-4{x}_{1}{x}_{2}}=sqrt{3}sqrt{{(frac{16}{11})}^{2}+frac{16}{11}}=frac{36}{11}.Also, the distance from point O to the line is d=frac{sqrt{2}}{sqrt{3}}=frac{sqrt{6}}{3}.Therefore, the area of triangle MON is:S=frac{1}{2}×frac{36}{11}×frac{sqrt{6}}{3}=boxed{frac{6sqrt{6}}{11}}

🧠:[<think>Okay, so I need to figure out in which quadrant the complex number z is located. The problem says z is equal to (2i)/(1 + i). Hmm, complex numbers can be tricky, especially when they're fractions like this. Let me think about how to approach this.First, I remember that when you have a complex number in the denominator, a common technique is to multiply the numerator and the denominator by the complex conjugate of the denominator. The complex conjugate of (1 + i) would be (1 - i), right? Because you just change the sign of the imaginary part. So if I do that, it should help eliminate the imaginary unit from the denominator.Let me write that out. So z = (2i)/(1 + i). Multiply numerator and denominator by (1 - i):z = [2i * (1 - i)] / [(1 + i)(1 - i)]Now, let me compute the numerator first. Multiply 2i by (1 - i):2i * 1 = 2i2i * (-i) = -2i²Wait, but i squared is -1, so -2i² is -2*(-1) = 2. So the numerator becomes 2i + 2. Which is 2 + 2i.Now the denominator: (1 + i)(1 - i). That's a difference of squares. So 1² - (i)² = 1 - i². Again, i² is -1, so 1 - (-1) = 1 + 1 = 2.So putting it all together, z = (2 + 2i)/2. Which simplifies to (2/2) + (2i)/2 = 1 + i.Wait, hold on, that's interesting. So z simplifies to 1 + i? Let me check that again to make sure I didn't make a mistake.Starting with z = (2i)/(1 + i). Multiply numerator and denominator by (1 - i):Numerator: 2i*(1 - i) = 2i - 2i² = 2i - 2(-1) = 2i + 2 = 2 + 2i.Denominator: (1 + i)(1 - i) = 1 - i + i - i² = 1 - i². Since i + (-i) cancels out. Then 1 - (-1) = 2.So z = (2 + 2i)/2 = 1 + i. Yes, that's correct. So z is 1 + i.Now, in the complex plane, the real part is 1 and the imaginary part is 1. So the point (1,1) is in the first quadrant, right? Because both the real and imaginary parts are positive. The first quadrant is where x > 0 and y > 0. So that would be answer A: First quadrant.Wait, but let me confirm. The problem might be trying to trick me. Did I do all the steps correctly?Alternative approach: Maybe I can represent the complex numbers in their polar forms and divide them. Let's see.First, 2i is 0 + 2i, so in polar form, that's 2(cos(π/2) + i sin(π/2)). The denominator 1 + i. The modulus of 1 + i is sqrt(1² + 1²) = sqrt(2), and the argument is arctan(1/1) = π/4. So in polar form, that's sqrt(2)(cos(π/4) + i sin(π/4)).So when dividing two complex numbers in polar form, we divide their moduli and subtract their arguments. So z would have modulus 2 / sqrt(2) = sqrt(2), and argument π/2 - π/4 = π/4.Therefore, z in polar form is sqrt(2)(cos(π/4) + i sin(π/4)), which corresponds to the rectangular form sqrt(2)*cos(π/4) + i sqrt(2)*sin(π/4). Since cos(π/4) and sin(π/4) are both sqrt(2)/2, multiplying by sqrt(2) gives (sqrt(2)*sqrt(2)/2) + i (sqrt(2)*sqrt(2)/2) = (2/2) + i (2/2) = 1 + i. So that's the same result as before. So z is indeed 1 + i, which is in the first quadrant. So answer A.Wait, but maybe there's another way to check. Let me plot it mentally. The complex plane has the real part on the x-axis and the imaginary part on the y-axis. So 1 + i would be at (1,1). The first quadrant is where both coordinates are positive. Second quadrant is negative x, positive y. Third is both negative. Fourth is positive x, negative y. So yes, (1,1) is in the first quadrant. Therefore, answer A.But hold on, the answer options given are A: First, B: Second, C: Third, D: Fourth. So the correct answer is A.But let me double-check the original problem. It says z = (2i)/(1 + i). If I made a mistake in the calculation steps, maybe I would get a different result. Let me verify again step by step.Original z = 2i / (1 + i). Multiply numerator and denominator by (1 - i):Numerator: 2i*(1 - i) = 2i - 2i^2. Since i^2 = -1, that becomes 2i - 2*(-1) = 2i + 2 = 2 + 2i.Denominator: (1 + i)(1 - i) = 1*(1) + 1*(-i) + i*(1) + i*(-i) = 1 - i + i - i^2. The -i and +i cancel, leaving 1 - (-1) = 2.Therefore, z = (2 + 2i)/2 = 1 + i. So yes, that's correct. So (1,1), which is first quadrant. So answer A.Alternatively, maybe someone might think that multiplying numerator and denominator by i would also work. Let me try that method.Multiply numerator and denominator by i: z = (2i * i) / [(1 + i) * i] = (2i^2) / (i + i^2). Since i^2 is -1, numerator is 2*(-1) = -2. Denominator is i + (-1) = -1 + i. So z = -2 / (-1 + i). Hmm, that seems more complicated. Maybe proceed further.Multiply numerator and denominator by (-1 - i):Numerator: -2*(-1 - i) = 2 + 2iDenominator: (-1 + i)(-1 - i) = (-1)(-1) + (-1)(-i) + i*(-1) + i*(-i) = 1 + i - i - i^2 = 1 - (-1) = 2.So z = (2 + 2i)/2 = 1 + i. Same result. So regardless of the method, z is 1 + i. So definitely first quadrant.Is there any chance I messed up the quadrants? Let's recall:- First quadrant: Real positive, Imaginary positive.- Second quadrant: Real negative, Imaginary positive.- Third quadrant: Real negative, Imaginary negative.- Fourth quadrant: Real positive, Imaginary negative.So (1,1) is first, (-1,1) is second, (-1,-1) third, (1,-1) fourth. So yes, 1 + i is first quadrant. Therefore, answer A.But wait, the problem is given in Chinese, but the options are labeled A to D, with A being first quadrant. So according to my calculations, A is the answer.Is there any way the answer could be different? Let me check once more.Alternative approach: Maybe plot the original numerator and denominator.Numerator: 2i is at (0,2). Denominator: 1 + i is at (1,1). When you divide two complex numbers, the result has a magnitude equal to the quotient of their magnitudes and an angle equal to the difference of their angles.So the magnitude of 2i is 2, and the magnitude of 1 + i is sqrt(1 + 1) = sqrt(2). Therefore, magnitude of z is 2 / sqrt(2) = sqrt(2). The angle of 2i is 90 degrees (π/2 radians), and the angle of 1 + i is 45 degrees (π/4 radians). Therefore, the angle of z is 90 - 45 = 45 degrees (π/4 radians). Therefore, z is sqrt(2) * (cos(45°) + i sin(45°)), which as we saw earlier, is 1 + i. So again, (1,1) in the complex plane, first quadrant. Answer A.Therefore, I think the answer is definitely A. First quadrant.Final Answerboxed{A}</think>Given ( z = frac{2i}{1+i} ), we need to determine the quadrant in which the complex number ( z ) is located in the complex plane.First, we multiply the numerator and the denominator by the complex conjugate of the denominator to simplify the expression:[z = frac{2i}{1+i} cdot frac{1-i}{1-i} = frac{2i(1-i)}{(1+i)(1-i)}]Calculating the numerator:[2i(1-i) = 2i - 2i^2 = 2i + 2 quad (text{since } i^2 = -1)]Calculating the denominator:[(1+i)(1-i) = 1 - i^2 = 1 - (-1) = 2]Thus, the simplified form of ( z ) is:[z = frac{2 + 2i}{2} = 1 + i]The complex number ( 1 + i ) has a real part of 1 and an imaginary part of 1. In the complex plane, the point (1, 1) is located in the first quadrant because both the real and imaginary parts are positive.Therefore, the answer is (boxed{A}).,<think>Okay, so I need to figure out which quadrant the complex number z is located in. The problem says z is equal to (2i) divided by (1 + i). Hmm, let me think. I remember that when dealing with complex numbers in the denominator, it's helpful to multiply the numerator and denominator by the conjugate of the denominator to rationalize it. The conjugate of 1 + i is 1 - i, right? So if I do that, I can simplify the expression and find the real and imaginary parts of z, which will tell me the quadrant.Let me write this out step by step. Starting with z = (2i)/(1 + i). To rationalize the denominator, I'll multiply both numerator and denominator by (1 - i):z = [2i * (1 - i)] / [(1 + i)(1 - i)].First, let me compute the denominator. Multiplying (1 + i)(1 - i) is a difference of squares, so it should be 1^2 - i^2. Since i^2 is -1, that becomes 1 - (-1) = 2. So the denominator is 2.Now the numerator: 2i*(1 - i) = 2i - 2i^2. Again, i^2 is -1, so this becomes 2i - 2*(-1) = 2i + 2. So the numerator is 2 + 2i.Putting it all together, z = (2 + 2i)/2 = (2/2) + (2i)/2 = 1 + i. Wait, that's interesting. So z simplifies to 1 + i. But hold on, 1 + i is a complex number with real part 1 and imaginary part 1, which would be in the first quadrant. But let me double-check my steps to make sure I didn't make a mistake.Starting over: z = (2i)/(1 + i). Multiply numerator and denominator by (1 - i):Numerator: 2i*(1 - i) = 2i - 2i^2. Since i^2 = -1, that becomes 2i + 2.Denominator: (1 + i)(1 - i) = 1 - i + i - i^2 = 1 - i^2 = 1 - (-1) = 2.So z = (2 + 2i)/2 = 1 + i. Yep, that seems right. So z is 1 + i, which has a positive real part and a positive imaginary part. That would place it in the first quadrant. But the answer choices are A: First quadrant, B: Second, etc. So the answer should be A.Wait, but let me verify once more. Sometimes when simplifying complex numbers, it's easy to mix up the signs. Let me check each step again.Multiplying numerator and denominator by (1 - i):Numerator: 2i*(1) = 2i and 2i*(-i) = -2i^2. Then, -2i^2 is +2 because i^2 is -1. So total numerator is 2i + 2, which is 2 + 2i.Denominator: (1 + i)(1 - i). Multiply using FOIL: 1*1 + 1*(-i) + i*1 + i*(-i) = 1 - i + i - i^2. The -i and +i cancel, leaving 1 - i^2. Since i^2 is -1, that becomes 1 - (-1) = 2. So denominator is 2. Dividing numerator by 2 gives 1 + i. So yes, z = 1 + i. Therefore, the real part is 1 (positive) and the imaginary part is 1 (positive), which is in the first quadrant. The answer should be A. But wait, the options given are A to D. Let me check the problem again.Wait, the original problem says z = (2i)/(1 + i). Then the options are quadrants. Hmm, but according to my calculation, it's 1 + i, so first quadrant, which is A. But I have a feeling maybe I made a mistake here because sometimes when I rationalize, signs can flip. Wait, let me check another way. Maybe plotting the original complex numbers.Alternatively, maybe convert both numerator and denominator to polar form and divide them? Let's see.The numerator is 2i, which is 0 + 2i. In polar form, that's 2(cos(π/2) + i sin(π/2)).The denominator is 1 + i. The magnitude of 1 + i is sqrt(1^2 + 1^2) = sqrt(2). The angle (argument) is arctan(1/1) = π/4. So in polar form, 1 + i is sqrt(2)(cos(π/4) + i sin(π/4)).So when we divide two complex numbers in polar form, we divide their magnitudes and subtract their angles. So z = (2 / sqrt(2)) * [cos(π/2 - π/4) + i sin(π/2 - π/4)].Simplify the magnitude: 2 / sqrt(2) = sqrt(2).The angle is π/2 - π/4 = π/4. So z = sqrt(2)(cos(π/4) + i sin(π/4)). Which is the same as sqrt(2)*e^{iπ/4}. But in rectangular form, cos(π/4) is sqrt(2)/2, so sqrt(2)*(sqrt(2)/2 + i sqrt(2)/2) = sqrt(2)*(sqrt(2)/2) + i sqrt(2)*(sqrt(2)/2) = (2/2) + i(2/2) = 1 + i. Yep, same result. So z is 1 + i, which is in the first quadrant. Therefore, the answer is A.But wait, hold on a second. Wait, the original problem is given as z = 2i/(1 + i). Let me check with substitution. Suppose I plug in 1 + i into the denominator. If z = (1 + i), then 2i/(1 + i) would be 2i*(1 - i)/2, which is (2i - 2i^2)/2 = (2i + 2)/2 = 1 + i. So that's correct. So z = 1 + i. So real part is 1, imaginary part is 1. So coordinates (1,1) in the complex plane, which is first quadrant. So answer is A.But I have a sense that I might be missing something here because sometimes problems like this might trick you. Wait, maybe I misapplied the conjugate? Wait, when you have a denominator a + bi, the conjugate is a - bi. So in this case, denominator is 1 + i, so conjugate is 1 - i. Multiplying numerator and denominator by that is correct.Alternatively, let me compute it step by step:Numerator: 2i*(1 - i) = 2i*1 - 2i*i = 2i - 2i^2 = 2i - 2*(-1) = 2i + 2.Denominator: (1 + i)(1 - i) = 1*1 + 1*(-i) + i*1 + i*(-i) = 1 - i + i - i^2 = 1 - i + i - (-1) = 1 + 1 = 2. So denominator is 2.Therefore, z = (2 + 2i)/2 = 1 + i. Yep, same result. So all methods confirm that z is 1 + i. Therefore, the real part is positive, imaginary part is positive, so first quadrant. Answer is A.But wait, the options given are A: First quadrant, B: Second quadrant, C: Third quadrant, D: Fourth quadrant. So the correct answer is A. Hmm, but why does the question even ask? It seems straightforward. Maybe I miscalculated.Wait, let me check another approach. Suppose I represent 1 + i as a vector in the complex plane. Then dividing 2i by 1 + i is equivalent to scaling and rotating the vector 2i by the reciprocal of 1 + i. The reciprocal of 1 + i is (1 - i)/2, as we saw earlier. So multiplying 2i by (1 - i)/2 gives (2i*(1 - i))/2 = (2i - 2i^2)/2 = (2i + 2)/2 = 1 + i. So same result. Alternatively, if I use the formula for division of complex numbers: if z = a + ib and w = c + id, then z/w = (ac + bd)/(c² + d²) + i(bc - ad)/(c² + d²). Let's apply this formula.Here, z = 0 + 2i, and w = 1 + i. So a = 0, b = 2, c = 1, d = 1.Then real part is (0*1 + 2*1)/(1² + 1²) = (0 + 2)/2 = 1.Imaginary part is (2*1 - 0*1)/2 = (2 - 0)/2 = 1.So z = 1 + i. Same answer. Therefore, definitely in the first quadrant. So answer is A. But wait, I feel like maybe the answer is supposed to be different? Let me check once more.Wait, maybe I made a mistake in the initial step. Let me verify:Original problem: z = (2i)/(1 + i). Multiply numerator and denominator by (1 - i):Numerator: 2i*(1 - i) = 2i - 2i^2. Since i^2 is -1, this is 2i + 2. So numerator is 2 + 2i.Denominator: (1 + i)(1 - i) = 1 - i + i - i^2 = 1 - i^2 = 1 - (-1) = 2. So denominator is 2.Therefore, z = (2 + 2i)/2 = 1 + i. So real part is 1, imaginary part is 1. So (1,1) in the complex plane, which is first quadrant. So answer is A. I think that's correct. Maybe the problem is straightforward and the answer is A. I don't see a mistake here.Alternatively, maybe the question had a typo, but assuming it's written correctly, the answer should be A. So the point corresponding to z is in the first quadrant.Final Answerboxed{A}</think>To determine the quadrant in which the complex number ( z = frac{2i}{1+i} ) is located, we start by rationalizing the denominator. We multiply both the numerator and the denominator by the conjugate of the denominator, ( 1 - i ):[z = frac{2i}{1+i} cdot frac{1-i}{1-i} = frac{2i(1-i)}{(1+i)(1-i)}]First, we compute the denominator:[(1+i)(1-i) = 1^2 - i^2 = 1 - (-1) = 2]Next, we compute the numerator:[2i(1-i) = 2i - 2i^2 = 2i - 2(-1) = 2i + 2]Putting these together, we get:[z = frac{2i + 2}{2} = frac{2 + 2i}{2} = 1 + i]The complex number ( 1 + i ) has a real part of 1 and an imaginary part of 1, both of which are positive. Therefore, the point corresponding to ( z ) in the complex plane is located in the first quadrant.[boxed{A}]]

🔑:To determine the quadrant of the complex number z=frac{2i}{1+i}, we first simplify z by multiplying both the numerator and the denominator by the conjugate of the denominator:[z = frac{2i}{1+i} = frac{2i(1-i)}{(1+i)(1-i)}.]Expanding the numerator and the denominator:[z = frac{2i - 2i^2}{1 - i^2}.]Since i^2 = -1, we can simplify further:[z = frac{2i - 2(-1)}{1 - (-1)} = frac{2i + 2}{2} = 1 + i.]The complex number 1 + i corresponds to the point (1, 1) in the complex plane. This point is located in the first quadrant because both the real part (1) and the imaginary part (1) are positive.Therefore, the correct answer is boxed{text{A}}.

🧠:[<think>Okay, so I need to find the limit as n approaches infinity of the sequence n√n minus the square root of n(n+1)(n+2). Hmm, let me write that down again to make sure I have it right:lim(n→∞) [n√n - √(n(n+1)(n+2))]Alright, first off, let's understand each term separately. The first term is n multiplied by the square root of n. That simplifies to n^(3/2), right? Because n is n^1 and √n is n^(1/2), so when you multiply them, you add the exponents: 1 + 1/2 = 3/2. So the first term is n^(3/2).The second term is the square root of n(n+1)(n+2). Let me see, inside the square root, we have a product of three consecutive terms: n, n+1, and n+2. So expanding that might be a bit messy, but maybe there's a smarter way to approximate it for large n. Since n is going to infinity, perhaps we can approximate n+1 ≈ n and n+2 ≈ n? But wait, that might be too crude. Alternatively, factor out n from each term.Let me try that. Inside the square root, n(n+1)(n+2) can be written as n*(n+1)*(n+2). Let's factor out n from each term. So:n(n+1)(n+2) = n * (n(1 + 1/n)) * (n(1 + 2/n)) = n * n * n * (1 + 1/n)(1 + 2/n) = n^3 * (1 + 1/n)(1 + 2/n)Therefore, the square root of that would be √[n^3 * (1 + 1/n)(1 + 2/n)] = √n^3 * √[(1 + 1/n)(1 + 2/n)] = n^(3/2) * √[(1 + 1/n)(1 + 2/n)]So the second term is n^(3/2) times the square root of (1 + 1/n)(1 + 2/n). Therefore, the original expression becomes:n^(3/2) - n^(3/2) * √[(1 + 1/n)(1 + 2/n)] = n^(3/2) [1 - √{(1 + 1/n)(1 + 2/n)}]So factoring out n^(3/2), the expression is n^(3/2) multiplied by [1 minus the square root term]. Now, perhaps I can expand the square root term using a Taylor series or binomial approximation for large n. Since 1/n is small when n is large, terms like 1/n and 2/n can be considered as small quantities.Let me denote x = 1/n. Then as n approaches infinity, x approaches 0. So, the expression inside the square root becomes (1 + x)(1 + 2x). Let's multiply that out:(1 + x)(1 + 2x) = 1 + 2x + x + 2x^2 = 1 + 3x + 2x^2So, √[1 + 3x + 2x^2]. Now, we can expand this square root using the Taylor expansion around x = 0. Remember that for small t, √(1 + t) ≈ 1 + t/2 - t^2/8 + ... So here, t = 3x + 2x^2. Let's plug that in:√[1 + 3x + 2x^2] ≈ 1 + (3x + 2x^2)/2 - ( (3x + 2x^2)^2 ) / 8 + ...But since x is 1/n, which is approaching zero, perhaps we can keep terms up to x^2, but maybe even x^1 is sufficient. Let's check.First, compute the first two terms:1 + (3x)/2 + (2x^2)/2 - [ (9x^2 + 12x^3 + 4x^4) ) /8 ] + higher order terms.Simplify:1 + (3/2)x + x^2 - (9x^2)/8 - (12x^3)/8 - ... Combine like terms. The x terms: (3/2)x. The x^2 terms: x^2 - 9x^2/8 = (8x^2 - 9x^2)/8 = (-x^2)/8. The x^3 terms: -12x^3/8 = -3x^3/2. So up to x^2, this is approximately:1 + (3/2)x - (1/8)x^2.But since x = 1/n, substituting back in:1 + (3/(2n)) - (1/(8n^2)).Therefore, the square root term is approximately 1 + 3/(2n) - 1/(8n^2) for large n.Therefore, our original expression becomes:n^(3/2) [1 - (1 + 3/(2n) - 1/(8n^2))] = n^(3/2) [ -3/(2n) + 1/(8n^2) ]Simplify this:n^(3/2) * (-3/(2n)) + n^(3/2) * (1/(8n^2)) = -3/(2n) * n^(3/2) + 1/(8n^2) * n^(3/2)Simplify each term:First term: -3/2 * n^(3/2 - 1) = -3/2 * n^(1/2)Second term: 1/8 * n^(3/2 - 2) = 1/8 * n^(-1/2) = 1/(8n^(1/2))So the entire expression is:- (3/2) √n + 1/(8√n )But as n approaches infinity, the first term is - (3/2) √n which goes to negative infinity, and the second term 1/(8√n) goes to zero. Wait, but this would suggest the limit is negative infinity, which contradicts with the original problem? Because if the first term is n√n and the second term is something slightly larger, maybe? Wait, let me check my approximation again.Wait, hold on. If we approximate the square root term as 1 + 3/(2n) - 1/(8n^2), then the expression becomes 1 - that, which is 1 - 1 - 3/(2n) + 1/(8n^2) = -3/(2n) + 1/(8n^2). Then multiplying by n^(3/2) gives -3/(2n) * n^(3/2) + 1/(8n^2)*n^(3/2) = -3/2 * n^(1/2) + 1/(8) * n^(-1/2). So indeed, as n approaches infinity, the first term dominates and goes to negative infinity. But the original expression is n√n - sqrt(n(n+1)(n+2)). If n√n is the first term, and the second term is sqrt(n(n+1)(n+2)), which is slightly larger than n√n, since the product n(n+1)(n+2) is n^3 + 3n^2 + 2n, so sqrt(n^3 + 3n^2 + 2n) is sqrt(n^3(1 + 3/n + 2/n^2)) = n^(3/2)sqrt(1 + 3/n + 2/n^2). So expanding that sqrt, as before, gives n^(3/2)(1 + (3/(2n)) + ...). Therefore, the second term is approximately n^(3/2) + (3/2)n^(1/2) + ... So subtracting that from n^(3/2) gives approximately - (3/2) sqrt(n) + ..., which would go to negative infinity. But that seems counterintuitive because the original problem is asking for the limit. But the problem says calculate the limit. If the answer is negative infinity, then that's the case, but maybe I made a mistake in the approximation?Wait, let me check again. Let me compute the original expression for n=1: 1*sqrt(1) - sqrt(1*2*3) = 1 - sqrt(6) ≈ 1 - 2.45 ≈ -1.45For n=2: 2*sqrt(2) - sqrt(2*3*4) ≈ 2.828 - sqrt(24) ≈ 2.828 - 4.899 ≈ -2.071For n=3: 3*sqrt(3) - sqrt(3*4*5) ≈ 5.196 - sqrt(60) ≈ 5.196 - 7.746 ≈ -2.55For n=10: 10*sqrt(10) ≈ 31.623; sqrt(10*11*12) = sqrt(1320) ≈ 36.332; difference ≈ -4.709n=100: 100*sqrt(100) = 100*10=1000; sqrt(100*101*102) = sqrt(100*10302) = sqrt(1030200) ≈ 1015. So difference ≈ 1000 - 1015 ≈ -15Wait, but according to the previous approximation, the difference should be about - (3/2) sqrt(n) + 1/(8 sqrt(n)). For n=100, that would be - (3/2)*10 + 1/(8*10) ≈ -15 + 0.0125 ≈ -14.9875, which is close to the actual difference of approximately -15. So that seems correct.So as n increases, the difference is going to -infinity. But the problem is to compute the limit. So if that's the case, the limit is -infinity. But wait, maybe I made a mistake in the problem statement?Wait, the problem says "Calculate the limit of the numerical sequence". So the expression is n√n - sqrt(n(n+1)(n+2)). As n increases, n√n is n^(3/2), and sqrt(n(n+1)(n+2)) is sqrt(n^3 + 3n^2 + 2n). Let's compare the leading terms:sqrt(n^3 + 3n^2 + 2n) = n^(3/2) sqrt(1 + 3/n + 2/n^2). Expanding this sqrt gives 1 + (3/(2n)) + ( (3/n)^2 / 8 ) + ... So sqrt(n(n+1)(n+2)) ≈ n^(3/2)(1 + 3/(2n) + ... ). Therefore, when you subtract sqrt(n(n+1)(n+2)) from n√n, you get:n^(3/2) - n^(3/2)(1 + 3/(2n) + ...) = - (3/2) n^(1/2) + ... So indeed, leading term is - (3/2) sqrt(n), which goes to -infinity as n approaches infinity.But the problem is presented as a limit, so maybe the answer is -infinity. However, I need to check if there's a different approach where maybe the limit is finite. Let me try another method.Alternatively, perhaps rationalizing the expression. Let me set a = n√n, b = sqrt(n(n+1)(n+2)), then the expression is a - b. To rationalize, multiply by (a + b)/(a + b):(a - b)(a + b)/(a + b) = (a² - b²)/(a + b)Compute a² and b²:a² = (n√n)² = n^3b² = [sqrt(n(n+1)(n+2))]^2 = n(n+1)(n+2) = n^3 + 3n^2 + 2nTherefore, a² - b² = n^3 - (n^3 + 3n^2 + 2n) = -3n^2 - 2nTherefore, the expression becomes (-3n^2 - 2n)/(a + b) = (-3n^2 - 2n)/(n√n + sqrt(n(n+1)(n+2)))Now, let's analyze the denominator. As n approaches infinity:n√n = n^(3/2)sqrt(n(n+1)(n+2)) ≈ sqrt(n^3 + 3n^2 + 2n) ≈ n^(3/2) + (3n^2)/(2n^(1/2)) ) by the expansion sqrt(n^3 + 3n^2 + ...) = n^(3/2)sqrt(1 + 3/n + ...) ≈ n^(3/2)(1 + 3/(2n)) so sqrt(n(n+1)(n+2)) ≈ n^(3/2) + (3/2)n^(1/2)Therefore, denominator ≈ n^(3/2) + n^(3/2) + (3/2)n^(1/2) = 2n^(3/2) + (3/2)n^(1/2)But wait, actually, each term in the denominator is approximately n^(3/2). So a + b ≈ n^(3/2) + n^(3/2) = 2n^(3/2). But the second term has a correction. Let me write denominator as:n√n + sqrt(n(n+1)(n+2)) = n^(3/2) + n^(3/2)(1 + 3/(2n) + ...) = n^(3/2) + n^(3/2) + (3/2)n^(1/2) + ... ≈ 2n^(3/2) + (3/2)n^(1/2)Therefore, the denominator is approximately 2n^(3/2), with a lower order term. So numerator is -3n^2 - 2n ≈ -3n^2. Therefore, the expression becomes approximately (-3n^2)/(2n^(3/2)) ) = (-3n^2)/(2n^(3/2)) ) = (-3/2) n^(1/2), which matches the previous result. Therefore, the limit is indeed -infinity.But wait, the problem says "numerical sequence". Maybe in some contexts, people consider sequences tending to negative infinity as having a limit, but depending on the convention, sometimes limit is said to not exist if it goes to infinity. However, in standard calculus, we do say the limit is negative infinity. But I need to confirm if that's acceptable here.Alternatively, maybe the problem has a typo, and the first term is supposed to be (n)th root of n, but that's written as n√n. But usually, nth root is written as n√... but in the problem statement, it's written as n sqrt{n}, which is n multiplied by sqrt(n). So I think the original problem is correct as stated.Alternatively, perhaps I made a mistake in the algebra. Let me check again.Original expression:n√n - sqrt(n(n+1)(n+2))Expressed as n^(3/2) - sqrt(n(n+1)(n+2))Expanding sqrt(n(n+1)(n+2)):As n(n+1)(n+2) = n^3 + 3n^2 + 2nSo sqrt(n^3 + 3n^2 + 2n) = sqrt(n^3(1 + 3/n + 2/n^2)) = n^(3/2)sqrt(1 + 3/n + 2/n^2)Using the expansion sqrt(1 + ε) ≈ 1 + ε/2 - ε²/8 for small ε:Here, ε = 3/n + 2/n^2. So sqrt(1 + 3/n + 2/n^2) ≈ 1 + (3/(2n) + 1/n^2) - ( (3/n + 2/n^2)^2 )/8But (3/n + 2/n^2)^2 = 9/n² + 12/n³ + 4/n^4, so divided by 8: 9/(8n²) + 3/(2n³) + 1/(2n^4). Therefore:sqrt(...) ≈ 1 + 3/(2n) + 1/n² - 9/(8n²) - 3/(2n³) - ...Simplify: 1 + 3/(2n) + (1 - 9/8)/n² + ... = 1 + 3/(2n) - 1/(8n²) + ...So that's the same as before. Then the expression becomes:n^(3/2) [1 - (1 + 3/(2n) - 1/(8n²) + ...)] = n^(3/2) [ -3/(2n) + 1/(8n²) + ... ] = -3/(2) * n^(1/2) + 1/(8) * n^(-1/2) + ...So yes, leading term is -3/2 sqrt(n), which tends to -infinity as n approaches infinity.Therefore, the limit is negative infinity. But wait, the problem might expect a finite limit. Maybe I need to re-express the original expression differently. Wait, is there a possibility that the problem was supposed to be written differently? For example, sometimes in these problems, the limit might look like it goes to infinity, but after proper manipulation, it turns out to be finite. However, in this case, according to my calculations, it's going to negative infinity. Let me check with another approach.Alternatively, factor out n from inside the square root:sqrt(n(n+1)(n+2)) = sqrt(n^3(1 + 1/n)(1 + 2/n)) = n^(3/2) sqrt((1 + 1/n)(1 + 2/n))So, original expression:n^(3/2) - n^(3/2) sqrt((1 + 1/n)(1 + 2/n)) = n^(3/2)[1 - sqrt((1 + 1/n)(1 + 2/n))]Let me denote the term inside the square root as (1 + a)(1 + b) where a = 1/n and b = 2/n. Then, sqrt((1 + a)(1 + b)) = sqrt(1 + a + b + ab) = sqrt(1 + 3/n + 2/n²). Which is the same as before.Using the expansion sqrt(1 + x) ≈ 1 + x/2 - x²/8 for small x:Here, x = 3/n + 2/n². So:sqrt(1 + 3/n + 2/n²) ≈ 1 + (3/(2n) + 1/n²) - ( (3/n + 2/n²)^2 )/8Which we already did, leading to the same result. So the expression is:n^(3/2)[ -3/(2n) + lower terms ] = -3/2 n^(1/2) + ... So limit is -infty.Alternatively, perhaps using the expansion sqrt(n(n+1)(n+2)) = n sqrt{(n+1)(n+2)} = n sqrt{n^2 + 3n + 2} = n sqrt{n^2(1 + 3/n + 2/n²)} = n * n sqrt{1 + 3/n + 2/n²} = n^2 sqrt{1 + 3/n + 2/n²}Then, sqrt{1 + 3/n + 2/n²} ≈ 1 + (3/(2n)) + ( (3/n)^2 /8 ) - ... So:sqrt(n(n+1)(n+2)) ≈ n^2 [1 + 3/(2n) + (9)/(8n²) - ...] = n^2 + (3/2)n + 9/8 - ...But wait, then the original expression is n^(3/2) - [n^2 + (3/2)n + 9/8 + ...]. Wait, but n^2 is much larger than n^(3/2) as n approaches infinity. So this would suggest that the expression is n^(3/2) - n^2 - ..., which tends to -infinity. But this contradicts the previous expression. Wait, but this method must be incorrect. Because sqrt(n(n+1)(n+2)) is sqrt(n^3 + 3n^2 + 2n), which is of order n^(3/2), not n^2. So my mistake here is in factoring. Let me check:sqrt(n(n+1)(n+2)) = sqrt{n * (n+1) * (n+2)}. If I factor n from each term, as I did earlier, it's n^3(1 + 1/n)(1 + 2/n). Then the square root is n^(3/2)sqrt{(1 + 1/n)(1 + 2/n)}. So the expression is n^(3/2) times something slightly larger than 1. Therefore, the original expression is n^(3/2) - [n^(3/2)(1 + 3/(2n) + ...)] = - (3/2)sqrt(n) + ..., which goes to -infty.Alternatively, if I factor n^2 inside the square root:Wait, n(n+1)(n+2) = n(n^2 + 3n + 2) = n^3 + 3n^2 + 2n. Taking square root gives sqrt(n^3 + 3n^2 + 2n). Let's factor n^3:sqrt(n^3(1 + 3/n + 2/n^2)) = n^(3/2)sqrt(1 + 3/n + 2/n^2). So same as before. So sqrt(n(n+1)(n+2)) = n^(3/2)sqrt(1 + 3/n + 2/n^2) ≈ n^(3/2)(1 + 3/(2n) + ...). Therefore, the difference is:n^(3/2) - n^(3/2)(1 + 3/(2n)) = - (3/2) n^(1/2) + ... which again tends to -infinity.So all approaches lead to the conclusion that the limit is -infinity. Therefore, unless there's a miscalculation I haven't noticed, the answer should be negative infinity.But let's check with another perspective. Let me set n = 1/t^2, so as n approaches infinity, t approaches 0 from the positive side. Then, substituting:Expression becomes:(1/t^2) * sqrt(1/t^2) - sqrt( (1/t^2)(1/t^2 + 1)(1/t^2 + 2) )Wait, that might complicate things. Alternatively, perhaps set m = sqrt(n), so n = m^2. Then as n approaches infinity, m approaches infinity. Then:Original expression becomes:m^2 * m - sqrt( m^2 (m^2 + 1)(m^2 + 2) ) = m^3 - sqrt{ m^2 (m^2 +1)(m^2 +2) }Simplify inside the sqrt:m^2*(m^2 +1)(m^2 +2) = m^2*(m^4 + 3m^2 + 2) = m^6 + 3m^4 + 2m^2So sqrt(m^6 + 3m^4 + 2m^2) = sqrt(m^6(1 + 3/m^2 + 2/m^4)) = m^3 sqrt(1 + 3/m^2 + 2/m^4) ≈ m^3(1 + 3/(2m^2) + ... )Therefore, the expression becomes:m^3 - [ m^3 + (3/2)m + ... ] = - (3/2)m + ... which is - (3/2) sqrt(n) + ... so again, as n approaches infinity, this tends to -infinity.So this also confirms the previous result.Alternatively, perhaps applying L’Hospital’s Rule? But this is a sequence, not a function, though we can consider it as a function of a real variable and apply L’Hospital. Let me try.Let me define f(n) = n√n - sqrt(n(n+1)(n+2)). Let me write this as n^(3/2) - sqrt(n^3 + 3n^2 + 2n). Let me set x = n, so x approaches infinity. So f(x) = x^(3/2) - sqrt(x^3 + 3x^2 + 2x). Let me consider f(x) as x approaches infinity.To apply L’Hospital’s Rule, we can write f(x) in a form suitable for the rule. However, L’Hospital applies to indeterminate forms like infinity - infinity, which is the case here. But to apply L’Hospital, we need to convert this into a fraction.Let me write f(x) = x^(3/2) - sqrt(x^3 + 3x^2 + 2x) = - [ sqrt(x^3 + 3x^2 + 2x) - x^(3/2) ]Let me set this as - [ sqrt(x^3 + 3x^2 + 2x) - x^(3/2) ]. Let me denote this as - [ A - B ], where A = sqrt(x^3 + 3x^2 + 2x), B = x^(3/2). Then, A - B = (A - B) * (A + B)/(A + B) ) = (A² - B²)/(A + B). We already did this earlier. As before, A² - B² = x^3 + 3x² + 2x - x^3 = 3x² + 2x. So A - B = (3x² + 2x)/(A + B). Therefore, f(x) = - (3x² + 2x)/(A + B)So f(x) = - (3x² + 2x)/(sqrt(x^3 + 3x^2 + 2x) + x^(3/2))Divide numerator and denominator by x^(3/2):Numerator: (3x² + 2x)/x^(3/2) = 3x^(1/2) + 2x^(-1/2)Denominator: [sqrt(x^3 + 3x² + 2x) + x^(3/2)] / x^(3/2) = sqrt(1 + 3/x + 2/x²) + 1Therefore, f(x) = - [3x^(1/2) + 2x^(-1/2)] / [sqrt(1 + 3/x + 2/x²) + 1]As x approaches infinity, sqrt(1 + 3/x + 2/x²) approaches 1, so denominator approaches 1 + 1 = 2. The numerator is 3 sqrt(x) + 2/sqrt(x), which approaches infinity. Therefore, the entire expression is - [ infinity ] / 2 = - infinity.Thus, confirming again the limit is -infinity.But maybe there's an error in the problem statement? Alternatively, perhaps the original problem had a different sign? For example, if it was sqrt(n(n+1)(n+2)) - n√n, then the limit would be +infinity. But as stated, it's the other way. The user's problem is as written: lim(n→∞) [n√n - sqrt(n(n+1)(n+2))].Alternatively, maybe the user made a typo and intended a different expression, but given the problem as stated, the limit is negative infinity. Therefore, unless there is a miscalculation in all these methods, which I can't find, the conclusion is that the limit is -infinity.But just to be thorough, let me compute one more term in the expansion to check if higher-order terms could affect the result. So going back to the expansion of sqrt(1 + 3/n + 2/n²). Let's use more terms. Let me write t = 3/n + 2/n². Then, sqrt(1 + t) ≈ 1 + t/2 - t²/8 + t^3/16 - ... So t = 3/n + 2/n². Then t² = (3/n)^2 + 2*(3/n)*(2/n²) + (2/n²)^2 = 9/n² + 12/n³ + 4/n^4. Similarly, t^3 = (3/n + 2/n²)^3 = 27/n³ + ... higher order terms.So sqrt(1 + t) ≈ 1 + (3/(2n) + 1/n²) - (9/(8n²) + 12/(8n³) + 4/(8n^4)) + (27/(16n³) + ... )Combine the terms:1 + 3/(2n) + 1/n² - 9/(8n²) - 12/(8n³) - 4/(8n^4) + 27/(16n³) + ... Simplify:1 + 3/(2n) + (1 - 9/8)/n² + (-12/8 + 27/16)/n³ + ... = 1 + 3/(2n) - 1/(8n²) + (-24/16 + 27/16)/n³ + ... = 1 + 3/(2n) - 1/(8n²) + 3/(16n³) + ... Thus, the expression inside the square root is approximated as 1 + 3/(2n) - 1/(8n²) + 3/(16n³) + ... So when we subtract this from 1, we get:1 - [1 + 3/(2n) - 1/(8n²) + 3/(16n³) + ... ] = -3/(2n) + 1/(8n²) - 3/(16n³) + ... Multiplying by n^(3/2):-3/(2n) * n^(3/2) + 1/(8n²) * n^(3/2) - 3/(16n³) * n^(3/2) + ... = -3/2 n^(1/2) + 1/(8) n^(-1/2) - 3/(16) n^(-3/2) + ... So as n approaches infinity, the dominant term is -3/2 sqrt(n), and the other terms vanish. Therefore, the limit is indeed -infinity.Therefore, after multiple approaches and verifications, I conclude that the limit is negative infinity.Final Answerboxed{-infty}</think>To calculate the limit of the numerical sequence:lim _{n rightarrow infty}(n sqrt{n}-sqrt{n(n+1)(n+2)})we start by analyzing each term separately. The first term ( n sqrt{n} ) simplifies to ( n^{3/2} ). The second term ( sqrt{n(n+1)(n+2)} ) can be approximated for large ( n ).First, we factor out ( n ) from each term inside the square root:sqrt{n(n+1)(n+2)} = sqrt{n^3 left(1 + frac{1}{n}right)left(1 + frac{2}{n}right)}Expanding the product inside the square root:sqrt{n^3 left(1 + frac{3}{n} + frac{2}{n^2}right)} = n^{3/2} sqrt{1 + frac{3}{n} + frac{2}{n^2}}Using the Taylor series expansion for ( sqrt{1 + epsilon} ) around ( epsilon = 0 ):sqrt{1 + frac{3}{n} + frac{2}{n^2}} approx 1 + frac{3}{2n} - frac{1}{8n^2} + cdotsThus, the second term approximates to:n^{3/2} left(1 + frac{3}{2n} - frac{1}{8n^2} + cdots right)Subtracting this from the first term ( n^{3/2} ):n^{3/2} - n^{3/2} left(1 + frac{3}{2n} - frac{1}{8n^2} + cdots right) = n^{3/2} left( -frac{3}{2n} + frac{1}{8n^2} - cdots right)Simplifying the expression:n^{3/2} left( -frac{3}{2n} right) + n^{3/2} left( frac{1}{8n^2} right) = -frac{3}{2} sqrt{n} + frac{1}{8} frac{1}{sqrt{n}} + cdotsAs ( n rightarrow infty ), the dominant term ( -frac{3}{2} sqrt{n} ) tends to negative infinity, while the other terms vanish. Therefore, the limit of the sequence is:boxed{-infty},<think>Okay, so I need to find the limit of the sequence n√n minus the square root of n(n+1)(n+2) as n approaches infinity. Hmm, let me try to break this down step by step. First, let's write out the expression again to make sure I have it right:lim_{n→∞} [n√n - √(n(n+1)(n+2))]Hmm, maybe I can simplify both terms separately first. Let's start with the first term, n√n. That's n multiplied by the square root of n, which is the same as n^(1) * n^(1/2) = n^(3/2). So the first term simplifies to n^(3/2). Now, the second term is the square root of the product of n, n+1, and n+2. Let's look at the expression inside the square root: n(n+1)(n+2). If I expand that, maybe I can find a way to approximate it for large n. Expanding it might be a bit tedious, but perhaps there's a smarter way. Alternatively, since n is approaching infinity, maybe I can approximate (n+1) and (n+2) as being approximately n for large n. But I need to be careful because even small differences can matter when multiplied together.Alternatively, maybe I can factor out n from each term inside the square root. Let's try that. So, n(n+1)(n+2) = n*(n+1)*(n+2). If I factor out n from each term, then n+1 = n(1 + 1/n), and n+2 = n(1 + 2/n). So substituting that in, we have:n * [n(1 + 1/n)] * [n(1 + 2/n)] = n * n * n * (1 + 1/n)(1 + 2/n) = n^3 * (1 + 1/n)(1 + 2/n)So inside the square root, we have n^3 multiplied by (1 + 1/n)(1 + 2/n). Then taking the square root of that would be sqrt(n^3) * sqrt((1 + 1/n)(1 + 2/n)). The sqrt(n^3) is n^(3/2), which is the same as the first term. So the second term is n^(3/2) multiplied by sqrt[(1 + 1/n)(1 + 2/n)]. Therefore, the entire expression becomes:n^(3/2) - n^(3/2) * sqrt[(1 + 1/n)(1 + 2/n)] = n^(3/2) [1 - sqrt{(1 + 1/n)(1 + 2/n)}]So factoring out n^(3/2), we have the expression as n^(3/2) multiplied by [1 - sqrt{(1 + 1/n)(1 + 2/n)}]. Now, perhaps we can expand the terms inside the square root using a Taylor series or binomial approximation for large n (since 1/n is small as n approaches infinity).Let me recall that for small x, sqrt(1 + x) ≈ 1 + x/2 - x²/8 + ... So maybe I can approximate the product (1 + 1/n)(1 + 2/n) first, then take the square root and expand it.First, let's compute (1 + 1/n)(1 + 2/n). Multiplying these two binomials:(1 + 1/n)(1 + 2/n) = 1*1 + 1*(2/n) + (1/n)*1 + (1/n)*(2/n) = 1 + 2/n + 1/n + 2/n² = 1 + 3/n + 2/n².So the product under the square root is 1 + 3/n + 2/n². Therefore, sqrt(1 + 3/n + 2/n²). Let me denote x = 3/n + 2/n². So for large n, x is small, so I can approximate sqrt(1 + x) ≈ 1 + x/2 - x²/8 + ... Let's compute x first: x = 3/n + 2/n². Then x/2 = (3/(2n)) + (1)/n². Then x² = (3/n + 2/n²)^2 = 9/n² + 12/n³ + 4/n^4. So x²/8 ≈ (9)/(8n²) + (12)/(8n³) + ... So up to the 1/n² terms, sqrt(1 + x) ≈ 1 + (3/(2n) + 1/n²) - (9)/(8n²) + ... Let's collect terms:1 + 3/(2n) + (1 - 9/8)/n² + ... = 1 + 3/(2n) - 1/(8n²) + ...Therefore, sqrt(1 + 3/n + 2/n²) ≈ 1 + 3/(2n) - 1/(8n²) + higher order terms.Therefore, substituting back into the original expression, we have:n^(3/2) [1 - (1 + 3/(2n) - 1/(8n²) + ...)] = n^(3/2) [ -3/(2n) + 1/(8n²) + ... ] = n^(3/2) * [ -3/(2n) + 1/(8n²) + ... ]Simplify term by term:First term: n^(3/2) * (-3/(2n)) = (-3/2) * n^(3/2 - 1) = (-3/2) * n^(1/2) = (-3/2)√nSecond term: n^(3/2) * (1/(8n²)) = (1/8) * n^(3/2 - 2) = (1/8) * n^(-1/2) = 1/(8√n)So the expression becomes (-3/2)√n + 1/(8√n) + ... as n approaches infinity. Then, taking the limit as n approaches infinity, the leading term is (-3/2)√n, which goes to negative infinity. But that contradicts the original problem because the limit is supposed to be finite? Wait, maybe I made a mistake in my approximation.Wait, let me double-check. The original expression is n√n - sqrt(n(n+1)(n+2)). If we approximate sqrt(n(n+1)(n+2)) as n^(3/2) (1 + 3/(2n) - 1/(8n²) + ...), then subtracting that from n^(3/2), which is the first term, gives:n^(3/2) - n^(3/2)(1 + 3/(2n) - 1/(8n²) + ...) = n^(3/2) [1 - (1 + 3/(2n) - 1/(8n²) + ...)] = n^(3/2) [ -3/(2n) + 1/(8n²) - ... ] = (-3/2)n^(1/2) + (1/8)n^(-1/2) - ... So, as n approaches infinity, the first term is dominant and is negative infinity. But this can't be correct because when we subtract sqrt(n(n+1)(n+2)) from n√n, both terms are positive, but perhaps the sqrt(n(n+1)(n+2)) is larger than n√n? Let me check with n=1:n=1: 1*1 - sqrt(1*2*3) = 1 - sqrt(6) ≈ 1 - 2.45 ≈ -1.45. So negative.n=2: 2*sqrt(2) - sqrt(2*3*4) ≈ 2.828 - sqrt(24) ≈ 2.828 - 4.899 ≈ -2.071n=3: 3*sqrt(3) - sqrt(3*4*5) ≈ 5.196 - sqrt(60) ≈ 5.196 - 7.746 ≈ -2.55So the sequence is negative and seems to be approaching negative infinity? But the problem asks to compute the limit. If the limit is negative infinity, then that's the answer. But maybe I made a mistake in my expansion?Wait, but the problem might have a typo or perhaps I misread it? Let me check again.The problem is: limit as n approaches infinity of (n√n - sqrt(n(n+1)(n+2))). Hmm. So as n increases, n√n is n^(3/2), and sqrt(n(n+1)(n+2)) is sqrt(n^3 + 3n^2 + 2n) ≈ sqrt(n^3 + 3n^2) ≈ n^(3/2) + (3n^2)/(2n^(3/2)) ) by binomial approximation. Wait, let me do that approach.Let me consider sqrt(n^3 + 3n^2 + 2n). Let me factor out n^3: sqrt(n^3(1 + 3/n + 2/n^2)) = n^(3/2)sqrt(1 + 3/n + 2/n^2). Then using the binomial approximation for sqrt(1 + x) ≈ 1 + x/2 - x^2/8 + ... where x = 3/n + 2/n^2. So sqrt(1 + x) ≈ 1 + (3/(2n) + 2/(2n^2)) - ( (3/n)^2 )/(8) + ... which is 1 + 3/(2n) + 1/n^2 - 9/(8n^2) + ... = 1 + 3/(2n) - (9/8 - 1)/n^2 + ... = 1 + 3/(2n) - 1/(8n^2) + ...Therefore, sqrt(n(n+1)(n+2)) ≈ n^(3/2)(1 + 3/(2n) - 1/(8n^2)). Therefore, n√n - sqrt(n(n+1)(n+2)) ≈ n^(3/2) - n^(3/2)(1 + 3/(2n) - 1/(8n^2)) = n^(3/2)( -3/(2n) + 1/(8n^2) ) = -3/(2) * n^(1/2) + 1/(8) n^(-1/2). As n approaches infinity, the term -3/2 * sqrt(n) dominates, so the limit is negative infinity. But the problem is presented as a limit of a numerical sequence. Is it possible that the answer is negative infinity? The problem didn't specify whether it's finite or not. But maybe I made a mistake in the approximation?Wait, perhaps there's a different approach. Let me consider rationalizing the expression. Let me denote a_n = n√n - sqrt(n(n+1)(n+2)). Multiply and divide by the conjugate:a_n = [n√n - sqrt(n(n+1)(n+2))] * [n√n + sqrt(n(n+1)(n+2))] / [n√n + sqrt(n(n+1)(n+2))]The numerator becomes (n√n)^2 - [sqrt(n(n+1)(n+2))]^2 = n^3 - n(n+1)(n+2)Let me compute that:n^3 - n(n+1)(n+2) = n^3 - [n(n^2 + 3n + 2)] = n^3 - n^3 - 3n^2 - 2n = -3n^2 - 2nSo the numerator is -3n^2 - 2n. The denominator is n√n + sqrt(n(n+1)(n+2)). Let's approximate the denominator for large n.First, sqrt(n(n+1)(n+2)) ≈ sqrt(n^3 + 3n^2 + 2n) ≈ n^(3/2) + (3n^2)/(2n^(3/2)) ) = n^(3/2) + (3/2) n^(1/2). Similarly, n√n = n^(3/2). Therefore, the denominator is approximately n^(3/2) + n^(3/2) + (3/2) n^(1/2) = 2n^(3/2) + (3/2) n^(1/2). For large n, the dominant term is 2n^(3/2). So the denominator ≈ 2n^(3/2).Therefore, the expression a_n ≈ (-3n^2 - 2n) / (2n^(3/2)) ) = (-3n^2)/(2n^(3/2)) ) - (2n)/(2n^(3/2)) ) = (-3/2) n^(1/2) - (1) n^(-1/2)Therefore, as n approaches infinity, a_n ≈ - (3/2) sqrt(n) - 1/sqrt(n) which tends to negative infinity. Therefore, the limit is negative infinity.But the problem is presented as a numerical sequence limit. In calculus, usually, if the limit goes to negative infinity, we just state that. However, sometimes problems might expect a finite limit. So maybe I made a mistake in my calculations. Let me check with another approach.Alternatively, let's expand sqrt(n(n+1)(n+2)) as sqrt(n^3 + 3n^2 + 2n). Let me write this as sqrt(n^3(1 + 3/n + 2/n^2)) = n^(3/2) sqrt(1 + 3/n + 2/n^2). Then, using the expansion sqrt(1 + x) ≈ 1 + x/2 - x²/8 + ..., where x = 3/n + 2/n². Therefore, sqrt(1 + 3/n + 2/n²) ≈ 1 + (3/(2n) + 1/n²) - (9/(8n²) + ...) as before. So sqrt(n(n+1)(n+2)) ≈ n^(3/2)(1 + 3/(2n) - 1/(8n²) + ...). Then subtracting this from n^(3/2):n^(3/2) - sqrt(n(n+1)(n+2)) ≈ n^(3/2) [1 - (1 + 3/(2n) - 1/(8n²))] ≈ n^(3/2) [ -3/(2n) + 1/(8n²) ] ≈ -3/(2) n^(1/2) + 1/(8) n^(-1/2). So as n approaches infinity, this expression tends to negative infinity. Therefore, the limit is negative infinity. However, in some contexts, people might consider the absolute value or the problem might have a typo. Alternatively, maybe I misread the original problem. Let me check again.Original problem: lim_{n→∞} [n√n - sqrt(n(n+1)(n+2))]Yes, that's correct. So according to all the approaches I tried, the limit is negative infinity. But let me test with actual large n.Take n = 1000:First term: 1000 * sqrt(1000) = 1000 * 31.6227766 ≈ 31622.7766Second term: sqrt(1000*1001*1002) ≈ sqrt(1000*1000*1000 + terms) ≈ sqrt(1.003006e9). Let's compute 1000*1001*1002:1000*1001 = 1,001,000; 1,001,000*1002 = 1,001,000*1000 + 1,001,000*2 = 1,001,000,000 + 2,002,000 = 1,003,002,000. So sqrt(1,003,002,000). Let's compute sqrt(1.003002e9). sqrt(1.003002e9) ≈ sqrt(1.003002)*sqrt(1e9) ≈ 1.0015 * 31622.7766 ≈ 31622.7766 + 0.0015*31622.7766 ≈ 31622.7766 + 47.434 ≈ 31670.2106Therefore, first term ≈ 31622.7766, second term ≈ 31670.2106. So the difference is 31622.7766 - 31670.2106 ≈ -47.434. But according to our previous expression, the leading term is -3/2 sqrt(n). For n=1000, sqrt(1000)=31.6227766, so -3/2 * 31.6227766 ≈ -47.434, which matches the actual calculation. So the difference is about -47.434, but what happens as n increases?If n=10,000:First term: 10,000 * 100 = 1,000,000. Wait, sqrt(10,000) = 100, so 10,000 * 100 = 1,000,000. Wait, no, sqrt(10,000) is 100. So n√n = 10,000 * 100 = 1,000,000.Second term: sqrt(10,000*10,001*10,002). Let's approximate:10,000*10,001 = 100,010,000; then times 10,002: 100,010,000*10,002 = 1,000,100,020,000. So sqrt(1,000,100,020,000). sqrt(1.00010002e12) ≈ 1,000,050.001. So the second term is approximately 1,000,050.001. Then the difference is 1,000,000 - 1,000,050.001 ≈ -50.001. According to the formula: -3/2 * sqrt(10,000) = -3/2 * 100 = -150. Wait, that's conflicting with the actual calculation. Wait, but sqrt(10,000*10,001*10,002) is sqrt(10,000^3 + ...) which is approximately 10,000^(3/2) + (3*10,000^2)/(2*10,000^(3/2)) ) = 10,000^(3/2) + (3/2)*10,000^(1/2). So 10,000^(3/2) is 10,000*100 = 1,000,000. Then the second term is 1,000,000 + (3/2)*100 = 1,000,000 + 150 = 1,000,150. Then the difference would be 1,000,000 - 1,000,150 = -150. But in reality, when I computed, it was approximately 1,000,050.001. There's a discrepancy here. So perhaps my approximation was too crude?Wait, let's compute sqrt(n(n+1)(n+2)) more accurately. For n=10,000:n=10,000, n+1=10,001, n+2=10,002.The product is 10,000*10,001*10,002. Let me compute this exactly:First, compute 10,000*10,001 = 100,010,000. Then multiply by 10,002:100,010,000 * 10,002 = 100,010,000 * 10,000 + 100,010,000 * 2 = 1,000,100,000,000 + 200,020,000 = 1,000,300,020,000.Then sqrt(1,000,300,020,000). Let's compute this square root. Let me note that (1,000,100)^2 = 1,000,200,010,000, which is less than 1,000,300,020,000. The difference is 1,000,300,020,000 - 1,000,200,010,000 = 100,010,000. So let's try (1,000,100 + x)^2 = 1,000,300,020,000.Expanding: (1,000,100)^2 + 2*1,000,100*x + x^2 = 1,000,300,020,000We know (1,000,100)^2 = 1,000,200,010,000. Then:1,000,200,010,000 + 2*1,000,100*x + x² = 1,000,300,020,000Subtract 1,000,200,010,000:2*1,000,100*x + x² = 100,010,000Assuming x is small compared to 1,000,100, we can ignore x²:2*1,000,100*x ≈ 100,010,000 → x ≈ 100,010,000 / (2*1,000,100) ≈ 100,010,000 / 2,000,200 ≈ 50.00249975So sqrt(1,000,300,020,000) ≈ 1,000,100 + 50.00249975 ≈ 1,000,150.0025Therefore, the second term is approximately 1,000,150.0025. The first term is 10,000*sqrt(10,000) = 10,000*100 = 1,000,000. Then the difference is 1,000,000 - 1,000,150.0025 ≈ -150.0025, which is approximately -150. So this aligns with our earlier approximation of -3/2 * sqrt(n) = -3/2 * 100 = -150. So why did my initial calculation for n=1000 give approximately -47.434, which is -3/2 * sqrt(1000) ≈ -3/2 * 31.6227766 ≈ -47.434. So seems consistent.So for n=1000, difference ≈ -47.434, for n=10,000, difference≈-150.0025, which is -3/2 * sqrt(n). So as n increases, the difference is becoming more negative without bound. Therefore, the limit is indeed negative infinity. However, in the problem statement, it just says "Calculate the limit of the numerical sequence". If the limit is negative infinity, then the answer is -∞. But maybe in the problem's context, they expect a different answer? Or perhaps there's a miscalculation.Wait, let me think again. Maybe I should approach this using asymptotic expansions. Let's denote the expression as:n√n - sqrt(n(n+1)(n+2)) = n^(3/2) - sqrt(n^3 + 3n^2 + 2n)Let me factor out n^3 inside the square root:sqrt(n^3 + 3n^2 + 2n) = n^(3/2)sqrt(1 + 3/n + 2/n^2)Then, using the expansion sqrt(1 + ε) ≈ 1 + ε/2 - ε²/8 + ..., where ε = 3/n + 2/n². So:sqrt(1 + 3/n + 2/n²) ≈ 1 + (3/(2n) + 1/n²) - ( (3/n)^2 + 2*(3/n)*(2/n²) + (2/n²)^2 ) /8 + ... Wait, actually, the expansion of sqrt(1 + ε) ≈ 1 + ε/2 - ε²/8 + ..., so ε = 3/n + 2/n², so ε² = (3/n + 2/n²)^2 = 9/n² + 12/n³ + 4/n^4.Therefore, sqrt(1 + ε) ≈ 1 + (3/(2n) + 2/(2n²)) - (9/n² + 12/n³ + 4/n^4)/8 + ... Simplify:1 + 3/(2n) + 1/n² - 9/(8n²) - 12/(8n³) - 4/(8n^4) + ... = 1 + 3/(2n) + (1 - 9/8)/n² - (3/2)/n³ - 1/(2n^4) + ... Which is 1 + 3/(2n) - 1/(8n²) - 3/(2n³) - ... Therefore, sqrt(n(n+1)(n+2)) ≈ n^(3/2)[1 + 3/(2n) - 1/(8n²) - ...] = n^(3/2) + (3/2)n^(1/2) - (1/8)n^(-1/2) - ... Therefore, the expression n^(3/2) - sqrt(n(n+1)(n+2)) ≈ n^(3/2) - [n^(3/2) + (3/2)n^(1/2) - (1/8)n^(-1/2) + ...] = - (3/2)n^(1/2) + (1/8)n^(-1/2) + ... Thus, as n→infty, this tends to -∞. Therefore, the limit is -∞. But the problem might expect a different approach. Alternatively, maybe the original problem had a plus sign instead of a minus sign? Because if it were plus, the limit might be finite. But as per the problem statement, it's a minus sign.Alternatively, perhaps there's an error in the problem. But assuming the problem is correct, the limit is -infinity. However, I should check once again with another method.Let me consider using the expansion of the product (n+1)(n+2). Let's expand (n+1)(n+2) = n² + 3n + 2. Therefore, n(n+1)(n+2) = n(n² + 3n + 2) = n³ + 3n² + 2n. Then, sqrt(n³ + 3n² + 2n). Let's factor n³:sqrt(n³(1 + 3/n + 2/n²)) = n^(3/2)sqrt(1 + 3/n + 2/n²). As before.Alternatively, write the original expression as:n^(3/2) - sqrt(n³ + 3n² + 2n) = n^(3/2) - sqrt(n³ + 3n² + 2n). Let me factor out n³ inside the square root:= n^(3/2) - n^(3/2)sqrt(1 + 3/n + 2/n²) = n^(3/2)[1 - sqrt(1 + 3/n + 2/n²)]Then, set h = 1/n. As n→∞, h→0. So the expression becomes:n^(3/2)[1 - sqrt(1 + 3h + 2h²)] = n^(3/2)[ - (3h + 2h²)/2 + ( (3h + 2h²)^2 )/8 - ... ] by expanding sqrt(1 + x) ≈ 1 + x/2 - x²/8 + ... for small x.Thus, x = 3h + 2h², so:sqrt(1 + x) ≈ 1 + (3h + 2h²)/2 - (9h² + 12h³ + 4h^4)/8 + ... ≈ 1 + (3/2)h + h² - (9/8)h² - (12/8)h³ - ... Therefore, 1 - sqrt(1 + x) ≈ - (3/2)h - h² + (9/8)h² + ... = - (3/2)h + (9/8 - 1)h² + ... = - (3/2)h + (1/8)h² + ... Therefore, the expression becomes:n^(3/2)[ - (3/2)h + (1/8)h² + ... ] = n^(3/2)[ - (3/2)(1/n) + (1/8)(1/n²) + ... ] = n^(3/2)[ -3/(2n) + 1/(8n²) + ... ] = -3/(2) n^(1/2) + 1/(8) n^(-1/2) + ... Again, leading to the same conclusion. So the limit is negative infinity.But the problem is from a calculus textbook or problem set? Usually, if the problem is given, the limit might be finite. So maybe there's a typo in the problem. Alternatively, maybe the original expression was intended to have a plus sign inside the square root? For example, sqrt(n(n+1)(n+2)) + n√n, but that seems unlikely. Alternatively, maybe the problem wants the absolute value, but the user wrote it correctly.Alternatively, maybe I made a miscalculation in my steps. Let me verify with another approach. Let's consider the expression:sqrt(n(n+1)(n+2)) = sqrt(n^3 + 3n^2 + 2n). Let's compare this with n^(3/2) + a n^(1/2) + b n^(-1/2) + ... for some constants a, b. Let me assume that sqrt(n^3 + 3n^2 + 2n) can be written as n^(3/2) + a n^(1/2) + b n^(-1/2) + ... Then squaring both sides:(n^(3/2) + a n^(1/2) + b n^(-1/2) + ...)^2 = n^3 + 3n^2 + 2n.Left side:n^3 + 2a n^(3/2 + 1/2) + (a^2 + 2b) n^(1 + 0) + ... = n^3 + 2a n^2 + (a^2 + 2b) n + ... Therefore, equate coefficients:n^3: 1 = 1, okay.n^2: 2a = 3 ⇒ a = 3/2.n term: a^2 + 2b = 2 ⇒ (9/4) + 2b = 2 ⇒ 2b = 2 - 9/4 = -1/4 ⇒ b = -1/8.Therefore, sqrt(n(n+1)(n+2)) ≈ n^(3/2) + (3/2) n^(1/2) - (1/8) n^(-1/2) + ... Therefore, the difference n^(3/2) - sqrt(...) ≈ n^(3/2) - [n^(3/2) + (3/2) n^(1/2) - (1/8) n^(-1/2)] = - (3/2) n^(1/2) + (1/8) n^(-1/2) + ... Thus, the leading term is - (3/2) sqrt(n), so the limit is indeed negative infinity.Therefore, after multiple approaches and verification, I conclude that the limit is negative infinity. So the answer is -∞. However, in the problem statement, it's presented as a numerical sequence. Depending on the conventions, sometimes in calculus, if the limit diverges to negative infinity, we write that the limit does not exist (in the finite sense), but in extended real numbers, it's -∞. Since the problem says "Calculate the limit", and in many contexts, it's acceptable to state the limit as -∞. Hence, the answer should be negative infinity.Final Answerboxed{-infty}</think>To find the limit of the sequence ( lim _{n rightarrow infty}(n sqrt{n}-sqrt{n(n+1)(n+2)}) ), we start by simplifying the terms separately.1. The first term ( n sqrt{n} ) simplifies to ( n^{3/2} ).2. The second term ( sqrt{n(n+1)(n+2)} ) needs to be approximated for large ( n ).Expanding the product inside the square root:[n(n+1)(n+2) = n^3 + 3n^2 + 2n]Factoring out ( n^3 ):[sqrt{n^3 + 3n^2 + 2n} = sqrt{n^3 left(1 + frac{3}{n} + frac{2}{n^2}right)} = n^{3/2} sqrt{1 + frac{3}{n} + frac{2}{n^2}}]Using the binomial approximation for ( sqrt{1 + x} ) where ( x = frac{3}{n} + frac{2}{n^2} ):[sqrt{1 + frac{3}{n} + frac{2}{n^2}} approx 1 + frac{3}{2n} - frac{1}{8n^2} + cdots]Thus,[sqrt{n(n+1)(n+2)} approx n^{3/2} left(1 + frac{3}{2n} - frac{1}{8n^2} + cdots right)]Substituting back into the original expression:[n^{3/2} - n^{3/2} left(1 + frac{3}{2n} - frac{1}{8n^2} + cdots right) = n^{3/2} left(1 - left(1 + frac{3}{2n} - frac{1}{8n^2} + cdots right)right)][= n^{3/2} left( - frac{3}{2n} + frac{1}{8n^2} - cdots right) = - frac{3}{2} n^{1/2} + frac{1}{8} n^{-1/2} + cdots]As ( n to infty ), the dominant term is ( - frac{3}{2} sqrt{n} ), which tends to ( -infty ).Therefore, the limit of the sequence is:[boxed{-infty}]]

🔑:## Solutionlim _{n rightarrow infty}(n sqrt{n}-sqrt{n(n+1)(n+2)})=begin{aligned}& =lim _{n rightarrow infty} frac{(n sqrt{n}-sqrt{n(n+1)(n+2)})(n sqrt{n}+sqrt{n(n+1)(n+2)})}{n sqrt{n}+sqrt{n(n+1)(n+2)}}= & =lim _{n rightarrow infty} frac{n^{3}-n(n+1)(n+2)}{n sqrt{n}+sqrt{n(n+1)(n+2)}}=lim _{n rightarrow infty} frac{n^{3}-nleft(n^{2}+n+2 n+2right)}{n sqrt{n}+sqrt{n(n+1)(n+2)}}= & =lim _{n rightarrow infty} frac{n^{3}-nleft(n^{2}+3 n+2right)}{n sqrt{n}+sqrt{n(n+1)(n+2)}}=lim _{n rightarrow infty} frac{n^{3}-n^{3}-3 n^{2}-2 n}{n sqrt{n}+sqrt{n(n+1)(n+2)}}= & =lim _{n rightarrow infty} frac{-3 n^{2}-2 n}{n sqrt{n}+sqrt{n(n+1)(n+2)}}=lim _{n rightarrow infty} frac{n^{-frac{3}{2}}left(-3 n^{2}-2 nright)}{n^{-frac{3}{2}}(n sqrt{n}+sqrt{n(n+1)(n+2))}}= & =lim _{n rightarrow infty} frac{-3 n^{frac{1}{2}}-2 n^{-frac{1}{2}}}{n^{-frac{3}{2}} n sqrt{n}+n^{-frac{3}{2}} sqrt{n(n+1)(n+2)}}= & =lim _{n rightarrow infty} frac{-3 sqrt{n}-frac{2}{sqrt{n}}}{1+sqrt{n^{-3} n(n+1)(n+2)}}=lim _{n rightarrow infty} frac{-3 sqrt{n}-frac{2}{sqrt{n}}}{1+sqrt{left(1+frac{1}{n}right)left(1+frac{2}{n}right)}}= & =left{frac{-infty-0}{1+sqrt{(1+0)(1+0)}=frac{-infty}{2}right}=-inftyend{aligned}## Problem Kuznetsov Limits 5-26

🧠:[<think>Okay, so I need to show that the sequence ( (x_n)_{n ge 1} ) is convergent and find its limit. The given conditions are that ( x_1 > 1 ) and for every natural number ( n ), the sum of the first ( n+1 ) terms equals the product of the first ( n+1 ) terms. That is,[x_1 + x_2 + cdots + x_{n+1} = x_1 x_2 cdots x_{n+1} quad forall n in mathbb{N}.]First, let me try to understand what this equation is saying. For each natural number ( n ), if I take the sum of the first ( n+1 ) terms of the sequence, it's equal to the product of those same terms. So, starting from ( n = 1 ), let's plug in some values to see if I can find a pattern or recurrence relation.For ( n = 1 ):The equation becomes the sum of the first 2 terms equals the product of the first 2 terms:[x_1 + x_2 = x_1 x_2.]Let me solve for ( x_2 ). Rearranging:[x_1 x_2 - x_2 = x_1 implies x_2(x_1 - 1) = x_1 implies x_2 = frac{x_1}{x_1 - 1}.]Okay, so ( x_2 ) is expressed in terms of ( x_1 ). Since ( x_1 > 1 ), the denominator ( x_1 - 1 ) is positive, so ( x_2 ) is positive as well.Now, let's check ( n = 2 ):The equation becomes the sum of the first 3 terms equals the product of the first 3 terms:[x_1 + x_2 + x_3 = x_1 x_2 x_3.]We already know ( x_1 ) and ( x_2 ) in terms of ( x_1 ). Let me substitute ( x_2 = frac{x_1}{x_1 - 1} ) into this equation:[x_1 + frac{x_1}{x_1 - 1} + x_3 = x_1 cdot frac{x_1}{x_1 - 1} cdot x_3.]Let me compute the left-hand side (LHS) and right-hand side (RHS) separately.First, LHS:[x_1 + frac{x_1}{x_1 - 1} + x_3 = x_1 + frac{x_1}{x_1 - 1} + x_3.]Let me combine the first two terms:[x_1 + frac{x_1}{x_1 - 1} = frac{x_1(x_1 - 1) + x_1}{x_1 - 1} = frac{x_1^2 - x_1 + x_1}{x_1 - 1} = frac{x_1^2}{x_1 - 1}.]So, LHS simplifies to ( frac{x_1^2}{x_1 - 1} + x_3 ).Now, RHS:[x_1 cdot frac{x_1}{x_1 - 1} cdot x_3 = frac{x_1^2}{x_1 - 1} cdot x_3.]So, the equation becomes:[frac{x_1^2}{x_1 - 1} + x_3 = frac{x_1^2}{x_1 - 1} cdot x_3.]Let me subtract ( frac{x_1^2}{x_1 - 1} ) from both sides:[x_3 = frac{x_1^2}{x_1 - 1} (x_3 - 1).]Then, divide both sides by ( x_3 - 1 ), assuming ( x_3 neq 1 ):[frac{x_3}{x_3 - 1} = frac{x_1^2}{x_1 - 1}.]Solving for ( x_3 ):Cross-multiplying:[x_3 (x_1 - 1) = x_1^2 (x_3 - 1).]Expanding both sides:Left: ( x_3 x_1 - x_3 )Right: ( x_1^2 x_3 - x_1^2 )Bring all terms to left side:[x_3 x_1 - x_3 - x_1^2 x_3 + x_1^2 = 0.]Factor out ( x_3 ):[x_3 (x_1 - 1 - x_1^2) + x_1^2 = 0.]Wait, maybe I can rearrange terms differently.Let me write:Left side after expansion: ( x_3 x_1 - x_3 )Right side: ( x_1^2 x_3 - x_1^2 )Bring right side to left:( x_3 x_1 - x_3 - x_1^2 x_3 + x_1^2 = 0 )Factor terms with ( x_3 ):( x_3 (x_1 - 1 - x_1^2) + x_1^2 = 0 )Then,( x_3 (-x_1^2 + x_1 - 1) + x_1^2 = 0 )Multiply both sides by -1:( x_3 (x_1^2 - x_1 + 1) - x_1^2 = 0 )Then,( x_3 (x_1^2 - x_1 + 1) = x_1^2 )Hence,( x_3 = frac{x_1^2}{x_1^2 - x_1 + 1} )So, ( x_3 ) is expressed in terms of ( x_1 ).Hmm. Let me note that ( x_1 > 1 ), so denominator ( x_1^2 - x_1 + 1 ). Since ( x_1^2 - x_1 + 1 ) is a quadratic in ( x_1 ), discriminant ( (-1)^2 - 4 times 1 times 1 = 1 - 4 = -3 < 0 ), so it's always positive. Therefore, ( x_3 ) is positive.Now, proceeding similarly, perhaps there's a pattern here. Let me check ( n = 3 ):For ( n = 3 ), the equation becomes the sum of the first 4 terms equals the product of the first 4 terms:[x_1 + x_2 + x_3 + x_4 = x_1 x_2 x_3 x_4.]Again, substitute the known expressions for ( x_2 ) and ( x_3 ):( x_2 = frac{x_1}{x_1 - 1} ), ( x_3 = frac{x_1^2}{x_1^2 - x_1 + 1} ).So, let's compute LHS and RHS.LHS:[x_1 + frac{x_1}{x_1 - 1} + frac{x_1^2}{x_1^2 - x_1 + 1} + x_4.]RHS:[x_1 cdot frac{x_1}{x_1 - 1} cdot frac{x_1^2}{x_1^2 - x_1 + 1} cdot x_4.]This is getting a bit complicated, but perhaps there's a telescoping pattern here. Let me compute the product first.Product terms:Multiply ( x_1 cdot frac{x_1}{x_1 - 1} = frac{x_1^2}{x_1 - 1} ), then multiply by ( frac{x_1^2}{x_1^2 - x_1 + 1} ):So, product is ( frac{x_1^2}{x_1 - 1} cdot frac{x_1^2}{x_1^2 - x_1 + 1} = frac{x_1^4}{(x_1 - 1)(x_1^2 - x_1 + 1)} ).Therefore, RHS is ( frac{x_1^4}{(x_1 - 1)(x_1^2 - x_1 + 1)} cdot x_4 ).Now, the LHS:First, compute the sum ( x_1 + x_2 + x_3 ):We already have from the previous steps:For ( n=1 ), sum ( x_1 + x_2 = x_1 x_2 ).For ( n=2 ), sum ( x_1 + x_2 + x_3 = x_1 x_2 x_3 ).Wait a second! Wait, that's an important observation. For each ( n ), the sum up to ( n+1 ) terms equals the product up to ( n+1 ) terms. Therefore, for ( n = 2 ), the sum ( x_1 + x_2 + x_3 = x_1 x_2 x_3 ). Therefore, in the case of ( n = 3 ), which requires the sum up to 4 terms equals the product up to 4 terms. But since the sum up to 3 terms is already equal to the product up to 3 terms, adding ( x_4 ) to both sides:Wait, let me think again.Wait, for each ( n ), the sum of the first ( n+1 ) terms equals the product of the first ( n+1 ) terms. So, when we go from ( n ) to ( n+1 ), we add a new term ( x_{n+2} ), and the sum becomes sum_{k=1}^{(n+1)+1} x_k = product_{k=1}^{(n+1)+1} x_k.But this seems recursive. Let me try to express the relation between ( x_{n+1} ) and the previous terms.Suppose that for a general ( n ), we have:[S_n = x_1 + x_2 + cdots + x_n = P_n = x_1 x_2 cdots x_n.]Wait, but according to the problem statement, it's for each ( n in mathbb{N} ), sum_{k=1}^{n+1} x_k = product_{k=1}^{n+1} x_k. So actually, the sum of ( n+1 ) terms equals the product of ( n+1 ) terms. Therefore, if we denote ( S_{n} = sum_{k=1}^{n} x_k ) and ( P_{n} = prod_{k=1}^{n} x_k ), then the condition is ( S_{n+1} = P_{n+1} ) for all ( n in mathbb{N} ).Therefore, for each ( n ge 1 ), ( S_{n+1} = P_{n+1} ).But note that ( S_{n+1} = S_n + x_{n+1} ), and ( P_{n+1} = P_n cdot x_{n+1} ).Therefore, we have:For each ( n ge 1 ),[S_n + x_{n+1} = P_n cdot x_{n+1}.]Which can be rewritten as:[S_n = x_{n+1} (P_n - 1).]But from the previous step, since ( S_n = P_{n} ) for ( n ge 2 )? Wait, no. Wait, for the original problem, for each ( n in mathbb{N} ), the sum up to ( n+1 ) terms equals the product up to ( n+1 ) terms. So, when ( n = 1 ), it's sum up to 2 terms equals product up to 2 terms. When ( n = 2 ), sum up to 3 terms equals product up to 3 terms, etc.So, in general, for each ( n ge 1 ), we have ( S_{n+1} = P_{n+1} ). Therefore, for each ( n ge 1 ), ( S_{n+1} = S_n + x_{n+1} = P_n x_{n+1} ).But from the previous step, ( S_{n} + x_{n+1} = P_n x_{n+1} ), so rearranged:( S_n = x_{n+1} (P_n - 1) ).But since ( S_{n} = P_{n} ) when considering ( n ) as the previous index. Wait, no. Wait, if ( S_{n+1} = P_{n+1} ), then for ( n' = n + 1 ), ( S_{n'} = P_{n'} ). But ( n' ) here would start from 2. Wait, maybe my confusion is arising from the indices.Let me try to reindex.Let’s denote for each ( m ge 2 ), ( S_m = P_m ). Then, for ( m ge 2 ), ( S_{m} = S_{m - 1} + x_{m} = P_{m - 1} cdot x_{m} ).Therefore, for each ( m ge 2 ), we have:[S_{m - 1} + x_{m} = P_{m - 1} cdot x_{m}.]Which implies:[S_{m - 1} = x_{m} (P_{m - 1} - 1).]But ( S_{m - 1} ) is the sum of the first ( m - 1 ) terms, and ( P_{m - 1} ) is the product of the first ( m - 1 ) terms. Therefore, if we denote ( m - 1 = k ), then for ( k ge 1 ):[S_k = x_{k + 1} (P_k - 1).]But we know that ( S_k = P_k ) when ( k + 1 ge 2 ), i.e., ( k ge 1 ). Wait, no. Wait, no, ( S_{k} ) is just the sum, not necessarily equal to the product. The equality ( S_{m} = P_{m} ) is given for each ( m ge 2 ), so when ( m = k + 1 ), then ( S_{k + 1} = P_{k + 1} ). Therefore, the recurrence comes from relating ( S_{k} ) to ( x_{k + 1} ).But maybe this approach is complicating things. Let's try to find a recursive formula for ( x_{n+1} ) in terms of previous terms.From the equation:( S_n + x_{n+1} = P_n cdot x_{n+1} ).Therefore,( S_n = x_{n+1} (P_n - 1) ).But ( S_n = sum_{k=1}^n x_k ), and ( P_n = prod_{k=1}^n x_k ). Therefore,( x_{n+1} = frac{S_n}{P_n - 1} ).This gives a recursive formula for ( x_{n+1} ) in terms of ( S_n ) and ( P_n ). But ( S_n = P_n ) when considering ( n ) as the index from the original problem. Wait, no. Wait, the original problem states that for all ( n in mathbb{N} ), ( S_{n+1} = P_{n+1} ). So, for each ( n ), ( S_{n+1} = P_{n+1} ). Therefore, when we write ( x_{n+1} = frac{S_n}{P_n - 1} ), this is valid for all ( n ge 1 ).But since ( S_{n+1} = S_n + x_{n+1} = P_{n+1} ), and ( P_{n+1} = P_n cdot x_{n+1} ), so combining these, we get:( S_n + x_{n+1} = P_n cdot x_{n+1} implies S_n = x_{n+1} (P_n - 1) implies x_{n+1} = frac{S_n}{P_n - 1} ).But ( S_n ) here is the sum of the first ( n ) terms. However, if we can express ( S_n ) in terms of ( P_n ), then perhaps we can find a relation.Wait, but from the previous index, when we look at ( n' = n - 1 ), then ( S_{n} = P_{n} ), because for ( n' = n - 1 ), the original condition gives ( S_{n' + 1} = P_{n' + 1} ), so ( S_n = P_n ).Wait, this is crucial. Let me clarify.Given that for each ( n in mathbb{N} ), ( S_{n + 1} = P_{n + 1} ).Therefore, for ( n = 1 ), we have ( S_2 = P_2 ).For ( n = 2 ), ( S_3 = P_3 ).For ( n = 3 ), ( S_4 = P_4 ), etc.But ( S_n = P_n ) only when ( n ge 2 ). For ( n = 1 ), ( S_1 = x_1 ), which is not necessarily equal to ( P_1 = x_1 ), but in this case, they are equal. Wait, ( S_1 = x_1 ), ( P_1 = x_1 ). So actually, even for ( n = 0 ), if we extend the definition, but the problem states ( n in mathbb{N} ), which usually starts at 1. However, in the problem statement, it's for all ( n in mathbb{N} ), which might be starting at 1, so ( n ge 1 ), hence ( n + 1 ge 2 ). So actually, for ( n ge 1 ), ( S_{n + 1} = P_{n + 1} ), which means ( S_2 = P_2 ), ( S_3 = P_3 ), etc., but ( S_1 = x_1 ) is not covered by the condition. So, S_1 is just x_1, and P_1 is also x_1. So, in that case, S_1 = P_1 trivially. But the problem states for all ( n in mathbb{N} ), so n starts at 1, hence the first equation is for n=1: S_2 = P_2, which is x1 + x2 = x1 x2.So, in general, for each ( n ge 1 ), S_{n + 1} = P_{n + 1}. Therefore, if we consider ( m = n + 1 ), then for each ( m ge 2 ), S_m = P_m. So, for all ( m ge 2 ), sum_{k=1}^m x_k = product_{k=1}^m x_k.But then, for each ( m ge 2 ), S_{m} = S_{m - 1} + x_m = P_{m - 1} x_m.But since S_{m} = P_{m}, we have:P_{m} = S_{m - 1} + x_m = P_{m - 1} x_m.But also, P_{m} = P_{m - 1} x_m.Therefore, combining these two equations:P_{m - 1} x_m = S_{m - 1} + x_m.Subtract x_m from both sides:P_{m - 1} x_m - x_m = S_{m - 1}x_m (P_{m - 1} - 1) = S_{m - 1}But S_{m - 1} = sum_{k=1}^{m - 1} x_k.But for ( m - 1 ge 1 ), since ( m ge 2 ), ( m - 1 ge 1 ). However, unless ( m - 1 ge 2 ), we cannot say that S_{m - 1} = P_{m - 1}. Wait, S_{m - 1} = P_{m - 1} only when ( m - 1 ge 2 ), i.e., ( m ge 3 ). Wait, no. Because the original condition is for all ( n in mathbb{N} ), S_{n + 1} = P_{n + 1}. So for m >= 2, S_m = P_m. Therefore, for m - 1 >= 2, i.e., m >= 3, S_{m - 1} = P_{m - 1}. But for m = 2, m - 1 = 1, which is S_1 = x_1, which is equal to P_1 = x_1. So in fact, S_{m - 1} = P_{m - 1} for all m >= 2.Therefore, the equation becomes:x_m (P_{m - 1} - 1) = P_{m - 1}Therefore,x_m = P_{m - 1} / (P_{m - 1} - 1)But since S_{m - 1} = P_{m - 1}, this gives:x_m = P_{m - 1} / (P_{m - 1} - 1) = 1 + 1/(P_{m - 1} - 1)But this is for all m >= 2.Therefore, we have a recursive formula for x_m in terms of P_{m - 1}.But since P_{m} = P_{m - 1} x_m, substitute x_m from above:P_{m} = P_{m - 1} * [P_{m - 1} / (P_{m - 1} - 1)] = P_{m - 1}^2 / (P_{m - 1} - 1)Hence, we get a recurrence relation for P_m:P_m = P_{m - 1}^2 / (P_{m - 1} - 1)Therefore, starting from P_1 = x_1, we can compute P_2, P_3, etc., using this recurrence.So, let me denote P_1 = x_1.P_2 = P_1^2 / (P_1 - 1)P_3 = P_2^2 / (P_2 - 1) = [ (P_1^2 / (P_1 - 1))^2 ] / [ (P_1^2 / (P_1 - 1)) - 1 ]Simplify denominator of P_3:Denominator: (P_1^2 / (P_1 - 1)) - 1 = [P_1^2 - (P_1 - 1)] / (P_1 - 1) = [P_1^2 - P_1 + 1] / (P_1 - 1)Therefore, P_3 = [ P_1^4 / (P_1 - 1)^2 ] / [ (P_1^2 - P_1 + 1)/(P_1 - 1) ) ] = [ P_1^4 / (P_1 - 1)^2 ] * [ (P_1 - 1) / (P_1^2 - P_1 + 1) ) ] = P_1^4 / [ (P_1 - 1)(P_1^2 - P_1 + 1) ) ]This is getting complicated, but perhaps we can see a pattern here.Alternatively, let me compute P_1, P_2, P_3 in terms of x_1:Given that P_1 = x_1P_2 = x_1^2 / (x_1 - 1)P_3 = [x_1^2 / (x_1 - 1)]^2 / ( [x_1^2 / (x_1 - 1)] - 1 )Let me compute denominator:[x_1^2 / (x_1 - 1)] - 1 = [x_1^2 - (x_1 - 1)] / (x_1 - 1) = [x_1^2 - x_1 + 1] / (x_1 - 1)So, P_3 = [x_1^4 / (x_1 - 1)^2] / [ (x_1^2 - x_1 + 1) / (x_1 - 1) ) ] = x_1^4 / [ (x_1 - 1)(x_1^2 - x_1 + 1) ) ]Similarly, P_4 would be P_3^2 / (P_3 - 1), and so on.It seems that each P_m is a rational function of x_1, but the denominator is growing in complexity. However, maybe if I can find a general formula for P_m, or notice that P_m follows a certain pattern.Alternatively, maybe instead of looking at P_m, we can look for a pattern in the terms x_n.From the previous calculations:x_2 = x_1 / (x_1 - 1)x_3 = x_1^2 / (x_1^2 - x_1 + 1)x_4 = ?Wait, let me compute x_4 using the formula x_{m} = P_{m - 1} / (P_{m - 1} - 1). For m = 4, x_4 = P_3 / (P_3 - 1)From above, P_3 = x_1^4 / [ (x_1 - 1)(x_1^2 - x_1 + 1) ) ]Therefore,x_4 = [ P_3 ] / (P_3 - 1 ) = [ x_1^4 / D ] / ( x_1^4 / D - 1 ) where D = (x_1 - 1)(x_1^2 - x_1 + 1)Simplify denominator:(x_1^4 - D) / D = (x_1^4 - (x_1 - 1)(x_1^2 - x_1 + 1)) / DCompute numerator:x_1^4 - (x_1 - 1)(x_1^2 - x_1 + 1)Expand the product:(x_1 - 1)(x_1^2 - x_1 + 1) = x_1^3 - x_1^2 + x_1 - x_1^2 + x_1 - 1 = x_1^3 - 2x_1^2 + 2x_1 - 1Therefore,x_1^4 - (x_1^3 - 2x_1^2 + 2x_1 - 1 ) = x_1^4 - x_1^3 + 2x_1^2 - 2x_1 + 1Therefore,x_4 = [ x_1^4 / D ] / [ (x_1^4 - x_1^3 + 2x_1^2 - 2x_1 + 1 ) / D ] = x_1^4 / (x_1^4 - x_1^3 + 2x_1^2 - 2x_1 + 1 )This is getting more complex, but perhaps if we factor the denominator:Denominator: x_1^4 - x_1^3 + 2x_1^2 - 2x_1 + 1.Not sure if this factors nicely. Let me try plugging in x_1 = 2 to see numerical values, maybe it can hint a pattern.Let me take x_1 = 2, which is greater than 1.Then,x_2 = 2 / (2 - 1) = 2x_3 = 2^2 / (2^2 - 2 + 1) = 4 / (4 - 2 + 1) = 4 / 3 ≈ 1.333...x_4 = 2^4 / (2^4 - 2^3 + 2*2^2 - 2*2 + 1 ) = 16 / (16 - 8 + 8 - 4 + 1) = 16 / (13) ≈ 1.2307...x_5 would be calculated using P_4, which is P_3 * x_4. Wait, P_3 was calculated as x_1^4 / [ (x_1 - 1)(x_1^2 - x_1 + 1) ) ]For x_1 = 2:P_3 = 16 / [ (1)(4 - 2 + 1) ) ] = 16 / (1 * 3) = 16/3 ≈ 5.333...Then, x_4 = P_3 / (P_3 - 1 ) = (16/3) / (16/3 - 1) = (16/3) / (13/3) ) = 16/13 ≈ 1.2307..., which matches earlier.Then, P_4 = P_3 * x_4 = (16/3) * (16/13) = 256 / 39 ≈ 6.564...Then, x_5 = P_4 / (P_4 - 1 ) = (256/39) / (256/39 - 1) = (256/39) / (217/39 ) = 256 / 217 ≈ 1.179...Similarly, x_6 would be P_5 / (P_5 - 1 ), etc.Observing the numerical values when x_1 = 2:x_1 = 2x_2 = 2x_3 ≈ 1.333...x_4 ≈ 1.2307...x_5 ≈ 1.179...x_6 ≈ ?Continuing:P_4 = 256 / 39 ≈ 6.5641x_5 ≈ 256/217 ≈ 1.179P_5 = P_4 * x_5 ≈ (256/39) * (256/217 ) = 65536 / 8463 ≈ 7.745Then x_6 = P_5 / (P_5 - 1 ) ≈ 7.745 / (7.745 - 1 ) ≈ 7.745 / 6.745 ≈ 1.148...Similarly, x_7 would be P_6 / (P_6 - 1 ), etc.So, the terms x_n are decreasing after x_2 (which was equal to x_1 = 2 in this case), approaching 1 from above. Wait, but x_1 = 2, x_2 = 2, x_3 ≈ 1.333, x_4 ≈ 1.2307, x_5 ≈ 1.179, x_6 ≈ 1.148, x_7 would be even closer to 1. So, it seems that the sequence is decreasing and approaching 1. Maybe the limit is 1?But wait, in the problem statement, we need to show that the sequence is convergent and find its limit. So perhaps the limit is 1. But we need to verify this.Alternatively, if I take another value for x_1, say x_1 = 3.Then,x_2 = 3 / (3 - 1 ) = 1.5x_3 = 3^2 / (9 - 3 + 1 ) = 9 / 7 ≈ 1.2857x_4 = 3^4 / [ (3 - 1)(9 - 3 + 1) ) ] = 81 / [ 2 * 7 ] = 81 / 14 ≈ 5.7857Wait, wait, no: x_4 = P_3 / (P_3 - 1 )Wait, first compute P_3.For x_1 = 3:P_1 = 3P_2 = 3^2 / (3 - 1 ) = 9 / 2 = 4.5P_3 = (4.5)^2 / (4.5 - 1 ) = 20.25 / 3.5 ≈ 5.7857Therefore, x_4 = P_3 / (P_3 - 1 ) ≈ 5.7857 / 4.7857 ≈ 1.209Wait, so x_4 ≈ 1.209, then P_4 = P_3 * x_4 ≈ 5.7857 * 1.209 ≈ 7.0Then x_5 = P_4 / (P_4 - 1 ) ≈ 7.0 / 6.0 ≈ 1.1666...Then P_5 ≈ 7.0 * 1.1666 ≈ 8.1666...x_6 ≈ 8.1666 / 7.1666 ≈ 1.14Similarly, x_7 ≈ ... approaching 1.So in this case, starting with x_1 = 3, the sequence x_n decreases towards 1 as well.So perhaps regardless of the initial value x_1 > 1, the sequence converges to 1. Therefore, the limit is 1.But we need to formalize this.First, we need to show that the sequence is convergent. To do this, we can try to show that it is decreasing and bounded below by 1, hence convergent by the Monotone Convergence Theorem.From the examples above, after the second term, the sequence seems to be decreasing. Let's check if this is generally true.First, note that x_1 > 1.x_2 = x_1 / (x_1 - 1 ). Since x_1 > 1, denominator x_1 - 1 > 0, so x_2 is positive. Let's compare x_2 with x_1:If x_1 > 1, then x_2 = x_1 / (x_1 - 1 ). Let's compute x_2 - x_1 = x_1 / (x_1 - 1 ) - x_1 = x_1 [ 1/(x_1 - 1 ) - 1 ] = x_1 [ (1 - (x_1 - 1 )) / (x_1 - 1 ) ] = x_1 [ (2 - x_1 ) / (x_1 - 1 ) ]Therefore, x_2 - x_1 = x_1 (2 - x_1 ) / (x_1 - 1 )So, the sign of x_2 - x_1 depends on (2 - x_1 ). If x_1 < 2, then 2 - x_1 > 0, so x_2 - x_1 > 0, hence x_2 > x_1.If x_1 = 2, then x_2 = 2 / (2 - 1 ) = 2, so x_2 = x_1.If x_1 > 2, then 2 - x_1 < 0, so x_2 - x_1 < 0, hence x_2 < x_1.So, depending on x_1, x_2 can be greater than, equal to, or less than x_1.But in the examples above, when x_1 = 2, x_2 = 2, then x_3 = 4/3 ≈ 1.333 < 2, and then the sequence continues decreasing. When x_1 = 3, x_2 = 1.5 < 3, so x_2 is already less than x_1, and then the sequence continues decreasing.Wait, so if x_1 > 2, x_2 < x_1. If x_1 = 2, x_2 = x_1. If 1 < x_1 < 2, x_2 > x_1.But in all cases, after x_2, the sequence seems to decrease towards 1. Let's see for the case when x_1 is between 1 and 2, say x_1 = 1.5.x_1 = 1.5x_2 = 1.5 / (1.5 - 1 ) = 1.5 / 0.5 = 3x_3 = (1.5)^2 / ( (1.5)^2 - 1.5 + 1 ) = 2.25 / (2.25 - 1.5 + 1 ) = 2.25 / 1.75 ≈ 1.2857x_4 = ?Compute P_3:P_1 = 1.5P_2 = 1.5 * 3 = 4.5P_3 = 4.5 * 1.2857 ≈ 5.7857Then x_4 = P_3 / (P_3 - 1 ) ≈ 5.7857 / 4.7857 ≈ 1.209Similarly, x_5 ≈ P_4 / (P_4 - 1 ), where P_4 ≈ 5.7857 * 1.209 ≈ 7.0, so x_5 ≈ 7 / 6 ≈ 1.1666...Continuing, x_6 ≈ 8.1666 / 7.1666 ≈ 1.14, etc., approaching 1.So even if x_1 is between 1 and 2, x_2 jumps above 2, then the subsequent terms decrease towards 1.Hence, regardless of x_1 > 1, the sequence seems to approach 1. So, the limit is likely 1.Now, to formalize this, we need to show two things:1. The sequence ( x_n ) is eventually decreasing and bounded below by 1, hence convergent.2. The limit satisfies the equation resulting from taking the limit on both sides of the recursive relation, which would lead to the limit being 1.First, let's try to show that the sequence is eventually decreasing and bounded below by 1.First, note that all terms x_n are greater than 1. Since x_1 > 1, and x_2 = x_1 / (x_1 - 1 ). If x_1 > 1, then x_1 - 1 > 0, so x_2 is positive.But is x_2 > 1?x_2 = x_1 / (x_1 - 1 ). Let's check if this is greater than 1:x_1 / (x_1 - 1 ) > 1Multiply both sides by (x_1 - 1 ), which is positive since x_1 > 1:x_1 > x_1 - 1Which simplifies to 0 > -1, which is always true. Therefore, x_2 > 1.Similarly, x_3 = P_2 / (P_2 - 1 ). Since P_2 = x_1 x_2 > 1 * 1 = 1, so P_2 > 1. Then, x_3 = P_2 / (P_2 - 1 ). Let's check if x_3 > 1:x_3 = P_2 / (P_2 - 1 ) > 1 ?Multiply both sides by (P_2 - 1 ) > 0:P_2 > P_2 - 1 => 0 > -1, true. So x_3 > 1.Similarly, by induction, assume that for some n >= 2, x_n > 1 and P_{n - 1} > 1. Then, P_n = P_{n - 1} x_n > 1 * 1 = 1. Then, x_{n + 1} = P_n / (P_n - 1 ) > 1, as before. Therefore, by induction, all x_n > 1.Hence, the sequence is bounded below by 1.Next, we need to show that the sequence is eventually decreasing.In the examples above, after x_2 (if x_1 <= 2) or starting from x_2 (if x_1 > 2), the sequence decreases. Let's check if this is generally true.Case 1: x_1 > 2.Then, x_2 = x_1 / (x_1 - 1 ) < x_1, as shown earlier. So, x_2 < x_1.Now, compute x_3 = P_2 / (P_2 - 1 ) = (x_1 x_2 ) / (x_1 x_2 - 1 )But x_1 x_2 = x_1 * [ x_1 / (x_1 - 1 ) ] = x_1^2 / (x_1 - 1 )So, x_3 = [ x_1^2 / (x_1 - 1 ) ] / [ x_1^2 / (x_1 - 1 ) - 1 ] = [ x_1^2 / (x_1 - 1 ) ] / [ (x_1^2 - (x_1 - 1 )) / (x_1 - 1 ) ] = x_1^2 / (x_1^2 - x_1 + 1 )Compare x_3 with x_2:x_2 = x_1 / (x_1 - 1 )x_3 = x_1^2 / (x_1^2 - x_1 + 1 )Compute x_3 - x_2:= x_1^2 / (x_1^2 - x_1 + 1 ) - x_1 / (x_1 - 1 )Let me compute this difference:Let me denote D1 = x_1^2 - x_1 + 1 and D2 = x_1 - 1.Then,x_3 - x_2 = (x_1^2 D2 - x_1 D1 ) / (D1 D2 )Compute numerator:x_1^2 (x_1 - 1 ) - x_1 (x_1^2 - x_1 + 1 )= x_1^3 - x_1^2 - x_1^3 + x_1^2 - x_1= (x_1^3 - x_1^3 ) + (-x_1^2 + x_1^2 ) + (-x_1 )= -x_1Therefore,x_3 - x_2 = -x_1 / (D1 D2 ) = -x_1 / [ (x_1^2 - x_1 + 1 )(x_1 - 1 ) ]Since x_1 > 2, denominator is positive, numerator is -x_1 < 0. Therefore, x_3 - x_2 < 0, so x_3 < x_2.Therefore, in this case, if x_1 > 2, then x_2 < x_1, x_3 < x_2, and so on. Hence, the sequence is decreasing from x_2 onwards.Similarly, for x_1 = 2, x_2 = 2, x_3 = 4/3 < 2, x_4 < x_3, etc., so decreasing from x_3 onwards.For the case when 1 < x_1 < 2, let's check whether the sequence becomes decreasing after some term.Take x_1 = 1.5.x_1 = 1.5, x_2 = 1.5 / 0.5 = 3.x_3 = (1.5 * 3 ) / (1.5 * 3 - 1 ) = 4.5 / 3.5 ≈ 1.2857 < 3.x_4 = (4.5 * 1.2857 ) / (4.5 * 1.2857 - 1 ) ≈ 5.7857 / 4.7857 ≈ 1.209 < 1.2857So, after x_2, the sequence decreases. Therefore, even if x_1 is between 1 and 2, x_2 is greater than x_1, but starting from x_3, the sequence decreases.Hence, in all cases, the sequence is eventually decreasing (from x_2 onwards if x_1 > 2, from x_3 onwards if 1 < x_1 <= 2 ), and bounded below by 1. Therefore, by the Monotone Convergence Theorem, the sequence converges.Let L be the limit. Since the sequence is convergent, let’s denote lim_{n→∞} x_n = L. Since all x_n > 1, L >= 1.We need to find L.To find L, consider the recursive relation:From earlier, for m >= 2,x_{m} = P_{m - 1} / (P_{m - 1} - 1 )But also, since P_{m} = P_{m - 1} x_{m }, and S_{m} = P_{m }.However, as m approaches infinity, if P_{m} converges to some limit, say Q, then Q would satisfy Q = Q * L, implying that either Q = 0 or L = 1. But since all P_{m} > 1, Q cannot be 0. Therefore, L must be 1.Alternatively, let's take the limit on both sides of the recursive formula:We have x_{n+1} = P_n / (P_n - 1 )Taking n → ∞, assuming both x_n and P_n converge:lim_{n→∞} x_{n+1} = lim_{n→∞} P_n / (P_n - 1 )Let L = lim_{n→∞} x_n = lim_{n→∞} x_{n+1}Also, since P_n = x_1 x_2 ... x_n, if the limit of x_n is L = 1, then the behavior of P_n depends on the convergence of the product. If x_n approaches 1, the product P_n may approach infinity if the convergence is too slow, but in our case, since the product is also related to the sum.But let's think differently. Since S_n = P_n for n >= 2. If the limit of x_n is L = 1, then as n increases, the sum S_n = x_1 + x_2 + ... + x_n ~ x_1 + x_2 + ... + x_{n} ≈ (n - k) * 1 + constants, since the first few terms are larger. But the product P_n = x_1 x_2 ... x_n. If the terms x_n approach 1, the product behaves like the product of terms approaching 1, but multiplied together many times. However, if the convergence of x_n to 1 is fast enough, the product might converge, but if not, it might diverge.However, in our case, for each n, S_n = P_n. If S_n ~ n (if x_n ~ 1 for large n), then P_n ~ n. But product P_n = x_1 x_2 ... x_n ~ n. However, for a product of numbers approaching 1, the product typically behaves exponentially, not linearly. Therefore, this suggests a contradiction unless L = 1 and the product somehow balances to grow linearly.Alternatively, maybe the convergence of x_n to 1 is such that the product P_n ~ n.But this line of reasoning is getting a bit tangled. Let's try to formalize the limit.Assuming that lim_{n→∞} x_n = L = 1, and lim_{n→∞} P_n = ∞ (since the product of terms greater than 1 diverges). However, from the relation S_n = P_n, which would mean that the sum S_n also tends to infinity, which is true because it's a sum of terms approaching 1, so S_n ~ n. But P_n ~ product of terms approaching 1, which can behave differently. However, if x_n = 1 + a_n, where a_n → 0, then log P_n = sum_{k=1}^n log(1 + a_k ) ≈ sum_{k=1}^n a_k, assuming a_k is small. Therefore, log P_n ≈ sum_{k=1}^n a_k. If P_n ~ n, then log P_n ~ log n, which would imply that sum_{k=1}^n a_k ~ log n. Therefore, a_k ~ 1/k, so that the sum diverges logarithmically. But in our case, S_n = P_n ~ n, so sum_{k=1}^n a_k ~ log n, but S_n = sum_{k=1}^n x_k = sum_{k=1}^n (1 + a_k ) = n + sum_{k=1}^n a_k ~ n + log n. But S_n = P_n ~ n, so n + log n ~ n, which implies that log n is negligible compared to n, which is true. Therefore, this suggests that if a_k ~ 1/k, then log P_n ~ log n, and P_n ~ e^{log n} = n, which matches S_n ~ n.Therefore, this heuristic suggests that x_n = 1 + a_n where a_n ~ 1/n, so that sum a_k ~ log n, and P_n ~ n. However, this is a bit beyond our needs.But returning to the original problem: we need to show that the sequence converges and find its limit. We have established that the sequence is bounded below by 1 and eventually decreasing, so it converges. Let L = lim_{n→∞} x_n.We need to find L. If we take the limit on both sides of the recursive formula:x_{n+1} = P_n / (P_n - 1 )As n → ∞, x_{n+1} → L, and P_n → ∞ (since x_n > 1 for all n, the product diverges). Therefore,L = lim_{n→∞} [ P_n / (P_n - 1 ) ] = lim_{n→∞} [ 1 / (1 - 1/P_n ) ] = 1 / (1 - 0 ) = 1.Therefore, the limit is 1.Hence, the sequence converges to 1.Final Answerboxed{1}</think>Given the sequence ( (x_n)_{n ge 1} ) with ( x_1 > 1 ) and satisfying the equation[x_1 + x_2 + cdots + x_{n+1} = x_1 x_2 cdots x_{n+1} quad forall n in mathbb{N},]we need to show that this sequence is convergent and find its limit.First, we observe that for each ( n in mathbb{N} ), the sum of the first ( n+1 ) terms equals the product of those terms. This leads to a recursive relation for the terms of the sequence. By examining specific cases and deriving the recursive formulas, we find:1. For ( n = 1 ), we have ( x_1 + x_2 = x_1 x_2 ), leading to ( x_2 = frac{x_1}{x_1 - 1} ).2. For ( n = 2 ), we derive ( x_3 = frac{x_1^2}{x_1^2 - x_1 + 1} ).3. Generalizing this, we find that each term ( x_{n+1} ) can be expressed in terms of the product of the previous terms, leading to the recurrence relation ( x_{n+1} = frac{P_n}{P_n - 1} ), where ( P_n = x_1 x_2 cdots x_n ).We show that all terms ( x_n ) are greater than 1 and that the sequence is eventually decreasing. By induction and analysis, we find that the sequence is bounded below by 1 and eventually decreasing, hence convergent by the Monotone Convergence Theorem.To find the limit, we consider the behavior of the product ( P_n ) and the sum ( S_n ). As ( n to infty ), the product ( P_n ) diverges to infinity, and the sum ( S_n ) also diverges. However, using the recursive relation and taking the limit, we find that the limit ( L ) must satisfy:[L = lim_{n to infty} frac{P_n}{P_n - 1} = 1.]Thus, the sequence converges to 1.[boxed{1}],<think>Okay, so I need to show that the sequence ( (x_n)_{n ge 1} ) is convergent and find its limit. The given conditions are that ( x_1 > 1 ) and for every natural number ( n ), the sum of the first ( n+1 ) terms equals the product of those terms. The equation is:[x_1 + x_2 + cdots + x_{n+1} = x_1 x_2 cdots x_{n+1} quad forall n in mathbb{N}]Hmm, let me try to unpack this. The equation holds for all natural numbers ( n ), which means that no matter how many terms we take (starting from 2 terms when ( n = 1 )), the sum equals the product. That's interesting. So each time we add a term, the equation still holds. Let me see if I can find a pattern or a recursive relationship here.First, let's look at the case when ( n = 1 ). Then, the equation becomes:[x_1 + x_2 = x_1 x_2]Given that ( x_1 > 1 ), we can solve for ( x_2 ):Let me rearrange the equation:[x_1 x_2 - x_2 = x_1 implies x_2 (x_1 - 1) = x_1 implies x_2 = frac{x_1}{x_1 - 1}]Okay, so ( x_2 ) is expressed in terms of ( x_1 ). Now, let's check the case for ( n = 2 ). Then, the equation becomes:[x_1 + x_2 + x_3 = x_1 x_2 x_3]We already know ( x_2 ) in terms of ( x_1 ), so let's substitute that in:First, compute ( x_1 + x_2 ). From the first equation, we know that sum is equal to ( x_1 x_2 ). So substituting into the second equation:[x_1 x_2 + x_3 = x_1 x_2 x_3]Again, let's solve for ( x_3 ):[x_1 x_2 x_3 - x_3 = x_1 x_2 implies x_3 (x_1 x_2 - 1) = x_1 x_2 implies x_3 = frac{x_1 x_2}{x_1 x_2 - 1}]But since ( x_1 x_2 = x_1 + x_2 ) from the first equation, let's substitute that in:[x_3 = frac{x_1 x_2}{(x_1 + x_2) - 1}]But wait, from ( x_1 + x_2 = x_1 x_2 ), so substituting:[x_3 = frac{x_1 x_2}{x_1 x_2 - 1} = frac{x_1 x_2}{(x_1 + x_2) - 1}]But since ( x_1 x_2 = x_1 + x_2 ), then:[x_3 = frac{x_1 x_2}{x_1 x_2 - 1} = frac{S}{S - 1}]Where ( S = x_1 x_2 ). But ( S = x_1 + x_2 ), so if I denote ( S_1 = x_1 ), ( S_2 = x_1 + x_2 ), then ( S_2 = x_1 x_2 ), and ( x_3 = S_2 / (S_2 - 1) ). Hmm, interesting. Let me see if this pattern continues.Suppose for the next term, when ( n = 3 ), the equation becomes:[x_1 + x_2 + x_3 + x_4 = x_1 x_2 x_3 x_4]Again, we can use the previous equation. From the case ( n = 2 ), we have:[x_1 + x_2 + x_3 = x_1 x_2 x_3]Therefore, substituting into the ( n = 3 ) equation:[(x_1 x_2 x_3) + x_4 = x_1 x_2 x_3 x_4]Solving for ( x_4 ):[x_1 x_2 x_3 x_4 - x_4 = x_1 x_2 x_3 implies x_4 (x_1 x_2 x_3 - 1) = x_1 x_2 x_3 implies x_4 = frac{x_1 x_2 x_3}{x_1 x_2 x_3 - 1}]Again, ( x_1 x_2 x_3 = x_1 + x_2 + x_3 ), so substituting that in:[x_4 = frac{x_1 x_2 x_3}{(x_1 + x_2 + x_3) - 1} = frac{S_3}{S_3 - 1}]Where ( S_3 = x_1 + x_2 + x_3 = x_1 x_2 x_3 ).So, it seems like each subsequent term ( x_{k+1} ) can be expressed as ( frac{S_k}{S_k - 1} ), where ( S_k = x_1 + x_2 + cdots + x_k = x_1 x_2 cdots x_k ).Therefore, this suggests a recursive formula:Given ( S_1 = x_1 ), then ( S_{k} = S_{k-1} + x_k ), but wait, actually, ( S_k = S_{k-1} + x_k ), but according to the given condition, ( S_{k} = x_1 x_2 cdots x_k ).Wait, no. Wait, actually, for each ( n ), ( S_{n} = x_1 + x_2 + cdots + x_{n} = x_1 x_2 cdots x_{n} ). Therefore, each ( S_n ) is both the sum and the product of the first ( n ) terms.Wait, but that can't be, unless for each ( n ), the sum of the first ( n ) terms equals their product. Wait, but the original problem says:"the sequence ( (x_n)_{n ge 1} ) having ( x_1 > 1 ) and satisfying the equation[x_1 + x_2 + cdots + x_{n+1} =x_1 x_2 cdots x_{n+1} ,quadforall ninmathbb{N} .]"Wait, so for all ( n in mathbb{N} ), the sum of the first ( n+1 ) terms equals the product of the first ( n+1 ) terms. So when ( n = 1 ), sum of first 2 terms equals product of first 2 terms. When ( n = 2 ), sum of first 3 terms equals product of first 3 terms, etc.Therefore, for each ( k ge 2 ), the sum of the first ( k ) terms equals the product of the first ( k ) terms. Therefore, we can say that for each ( k ge 2 ):[S_k = P_k]where ( S_k = x_1 + x_2 + cdots + x_k ) and ( P_k = x_1 x_2 cdots x_k ).Therefore, given that for each ( k ge 2 ), ( S_k = P_k ).Given that, perhaps we can find a recursive formula for ( x_{k} ).Let me see. For ( k ge 2 ):[S_k = S_{k-1} + x_k = P_{k-1} + x_k = P_k = P_{k-1} cdot x_k]Therefore:[P_{k-1} + x_k = P_{k-1} cdot x_k]Let's write this equation:[P_{k-1} cdot x_k - x_k = P_{k-1} implies x_k (P_{k-1} - 1) = P_{k-1} implies x_k = frac{P_{k-1}}{P_{k-1} - 1}]But since ( P_{k-1} = S_{k-1} ), which is the sum of the first ( k-1 ) terms, which is equal to the product of the first ( k-1 ) terms.Therefore, we can write:[x_k = frac{S_{k-1}}{S_{k-1} - 1}]But ( S_{k-1} = P_{k-1} ), so:[x_k = frac{P_{k-1}}{P_{k-1} - 1}]But then, since ( P_{k-1} = S_{k-1} ), which is equal to ( S_{k-2} + x_{k-1} ). Wait, maybe there is a way to express ( x_k ) in terms of ( x_{k-1} ).Alternatively, let's try to compute the first few terms to see a pattern.Given ( x_1 > 1 ).Compute ( x_2 ):From ( n = 1 ):( x_1 + x_2 = x_1 x_2 implies x_2 = frac{x_1}{x_1 - 1} )Compute ( x_3 ):From ( n = 2 ):( x_1 + x_2 + x_3 = x_1 x_2 x_3 )But we already know ( x_1 + x_2 = x_1 x_2 ), so substituting:( x_1 x_2 + x_3 = x_1 x_2 x_3 implies x_3 (x_1 x_2 - 1) = x_1 x_2 implies x_3 = frac{x_1 x_2}{x_1 x_2 - 1} )But ( x_1 x_2 = x_1 + x_2 ), so substitute:( x_3 = frac{x_1 + x_2}{(x_1 + x_2) - 1} )But ( x_1 + x_2 = x_1 x_2 ), so:( x_3 = frac{x_1 x_2}{x_1 x_2 - 1} )Similarly, ( x_4 ):From ( n = 3 ):( x_1 + x_2 + x_3 + x_4 = x_1 x_2 x_3 x_4 )But the sum ( x_1 + x_2 + x_3 = x_1 x_2 x_3 ), so substituting:( x_1 x_2 x_3 + x_4 = x_1 x_2 x_3 x_4 implies x_4 (x_1 x_2 x_3 - 1) = x_1 x_2 x_3 implies x_4 = frac{x_1 x_2 x_3}{x_1 x_2 x_3 - 1} )Again, since ( x_1 x_2 x_3 = x_1 + x_2 + x_3 ), substitute:( x_4 = frac{x_1 + x_2 + x_3}{(x_1 + x_2 + x_3) - 1} = frac{x_1 x_2 x_3}{x_1 x_2 x_3 - 1} )Therefore, it's clear that each term ( x_{k} ) can be written as:[x_{k} = frac{P_{k-1}}{P_{k-1} - 1}]Where ( P_{k-1} = x_1 x_2 cdots x_{k-1} = S_{k-1} ).So, if I denote ( P_{k} = S_{k} ), then:[x_{k} = frac{S_{k-1}}{S_{k-1} - 1}]But ( S_{k} = S_{k-1} + x_k ), and since ( S_{k} = P_{k} = P_{k-1} cdot x_k ), we have:[S_{k-1} + x_k = P_{k-1} cdot x_k]But from the expression for ( x_k ), ( x_k = frac{S_{k-1}}{S_{k-1} - 1} ), so substituting into ( S_{k} ):[S_{k} = S_{k-1} + frac{S_{k-1}}{S_{k-1} - 1} = S_{k-1} left( 1 + frac{1}{S_{k-1} - 1} right ) = S_{k-1} cdot frac{S_{k-1} - 1 + 1}{S_{k-1} - 1} = S_{k-1} cdot frac{S_{k-1}}{S_{k-1} - 1}]But also, ( S_{k} = P_{k} = P_{k-1} cdot x_k = S_{k-1} cdot x_k = S_{k-1} cdot frac{S_{k-1}}{S_{k-1} - 1} ), which matches. So that's consistent.Therefore, the sequence ( S_k ) (which is both the sum and the product of the first ( k ) terms) satisfies the recursion:[S_k = S_{k-1} cdot frac{S_{k-1}}{S_{k-1} - 1}]Therefore, we can model the behavior of ( S_k ) and then relate that back to the terms ( x_k ).Given that ( x_1 > 1 ), let's compute ( S_1 = x_1 > 1 ), then ( S_2 = x_1 x_2 = x_1 + x_2 ), which we already saw is equal to ( S_1 cdot frac{S_1}{S_1 - 1} ). Wait, let's check:From the recursion, ( S_2 = S_1 cdot frac{S_1}{S_1 - 1} ). Since ( S_1 = x_1 ), that gives ( S_2 = x_1 cdot frac{x_1}{x_1 - 1} ). But from the definition, ( S_2 = x_1 + x_2 = x_1 + frac{x_1}{x_1 - 1} ). Let me check if these are equal.Compute ( x_1 + frac{x_1}{x_1 - 1} = x_1 cdot left( 1 + frac{1}{x_1 - 1} right ) = x_1 cdot frac{x_1 - 1 + 1}{x_1 - 1} = x_1 cdot frac{x_1}{x_1 - 1} ). Yes, that's equal to ( S_2 = x_1 cdot frac{x_1}{x_1 - 1} ). So the recursion holds.Therefore, ( S_k = S_{k-1} cdot frac{S_{k-1}}{S_{k-1} - 1} ).So, this is a recursive sequence starting from ( S_1 = x_1 > 1 ), and each subsequent term is defined by ( S_k = frac{(S_{k-1})^2}{S_{k-1} - 1} ).So, we can analyze the behavior of ( S_k ). Let's write the recursion as:[S_k = frac{S_{k-1}^2}{S_{k-1} - 1}]Our goal is to find the limit of ( x_n ). Since ( x_k = frac{S_{k-1}}{S_{k-1} - 1} ), if we can find the limit of ( S_k ), then we can find the limit of ( x_k ).Assume that ( S_k ) converges to some limit ( L ). Then, taking limits on both sides of the recursion:[L = frac{L^2}{L - 1}]Multiply both sides by ( L - 1 ):[L(L - 1) = L^2 implies L^2 - L = L^2 implies -L = 0 implies L = 0]But ( S_1 = x_1 > 1 ), and let's check the behavior of ( S_k ):Given ( S_{k} = frac{S_{k-1}^2}{S_{k-1} - 1} ), let's compute the next few terms to see if they are increasing or decreasing.Suppose ( S_1 = x_1 = 2 ). Then,( S_2 = frac{2^2}{2 - 1} = 4 / 1 = 4 )( S_3 = frac{4^2}{4 - 1} = 16 / 3 ≈ 5.333 )( S_4 = frac{(16/3)^2}{(16/3) - 1} = (256/9) / (13/3) = (256/9) * (3/13) = 768 / 117 ≈ 6.564 )Similarly, ( S_4 ≈ 6.564 ), ( S_5 ≈ (6.564)^2 / (6.564 - 1) ≈ 43.09 / 5.564 ≈ 7.743 ), and so on. Each term seems to be increasing.Wait, but the fixed point equation gives L=0, which contradicts the increasing sequence starting from S1=2>1. Therefore, our assumption that S_k converges is likely wrong. Wait, but then how can the problem say to show that the sequence is convergent? Maybe there's a mistake in my analysis.Wait, perhaps S_k diverges to infinity? Let's see.Suppose ( S_{k} ) is increasing and tends to infinity. Then, as ( S_{k-1} ) approaches infinity, the recursion becomes:( S_k ≈ frac{S_{k-1}^2}{S_{k-1}} = S_{k-1} ), which would suggest that the growth slows down. But in reality, ( S_k = S_{k-1}^2 / (S_{k-1} - 1) ≈ S_{k-1} + 1 + o(1) ) as ( S_{k-1} to infty ). Wait, let's see:Express ( S_{k} = frac{S_{k-1}^2}{S_{k-1} - 1} = S_{k-1} + frac{S_{k-1}}{S_{k-1} - 1} ). Wait, that comes from:[frac{S_{k-1}^2}{S_{k-1} - 1} = S_{k-1} + frac{S_{k-1}}{S_{k-1} - 1}]Because:[frac{S_{k-1}^2}{S_{k-1} - 1} = frac{S_{k-1}(S_{k-1} - 1) + S_{k-1}}}{S_{k-1} - 1} = S_{k-1} + frac{S_{k-1}}{S_{k-1} - 1}]So, ( S_k = S_{k-1} + frac{S_{k-1}}{S_{k-1} - 1} )Therefore, since ( S_{k-1} > 1 ), the term ( frac{S_{k-1}}{S_{k-1} - 1} > 1 ), so each term ( S_k ) is greater than ( S_{k-1} + 1 ). Therefore, ( S_k ) increases at least linearly, but probably faster. Wait, but as ( S_{k-1} ) increases, the increment ( frac{S_{k-1}}{S_{k-1} - 1} ) approaches 1. So, when ( S_{k-1} ) is large, ( S_k approx S_{k-1} + 1 ), so approximately linear growth. But even linear growth would mean ( S_k to infty ). So if ( S_k ) tends to infinity, then what happens to ( x_k )?Given that ( x_{k} = frac{S_{k-1}}{S_{k-1} - 1} ), as ( S_{k-1} to infty ), ( x_{k} to frac{infty}{infty - 1} approx 1 ). So, ( x_k ) approaches 1 from above, since ( S_{k-1} > 1 ), so ( S_{k-1} - 1 > 0 ), and ( x_k = frac{S_{k-1}}{S_{k-1} - 1} = 1 + frac{1}{S_{k-1} - 1} ), so as ( S_{k-1} to infty ), ( x_k to 1 + 0 = 1 ).Therefore, the sequence ( x_k ) approaches 1. Therefore, the limit is 1. So, even though ( S_k ) tends to infinity, the terms ( x_k ) approach 1.But the problem states to show that the sequence ( x_n ) is convergent and find its limit. Therefore, the limit is 1. But we need to formalize this.First, let's show that ( S_k ) is strictly increasing and tends to infinity.We can prove by induction that ( S_k > S_{k-1} ). Base case: ( S_1 = x_1 > 1 ). Then, ( S_2 = frac{S_1^2}{S_1 - 1} ). Since ( S_1 > 1 ), ( S_1 - 1 > 0 ), so ( S_2 > 0 ). Now, compute ( S_2 - S_1 = frac{S_1^2}{S_1 - 1} - S_1 = frac{S_1^2 - S_1(S_1 - 1)}{S_1 - 1} = frac{S_1^2 - S_1^2 + S_1}{S_1 - 1} = frac{S_1}{S_1 - 1} > 0 ). So, ( S_2 > S_1 ).Assume ( S_k > S_{k-1} ). Then, ( S_{k+1} = frac{S_k^2}{S_k - 1} ). Then, ( S_{k+1} - S_k = frac{S_k^2}{S_k - 1} - S_k = frac{S_k^2 - S_k(S_k - 1)}{S_k - 1} = frac{S_k^2 - S_k^2 + S_k}{S_k - 1} = frac{S_k}{S_k - 1} > 0 ). Therefore, ( S_{k+1} > S_k ). Therefore, by induction, ( S_k ) is strictly increasing.Now, to show ( S_k to infty ). Suppose, for contradiction, that ( S_k ) converges to some finite limit ( L geq S_1 > 1 ). Then, taking the limit in the recursion:[L = frac{L^2}{L - 1} implies L(L - 1) = L^2 implies L^2 - L = L^2 implies -L = 0 implies L = 0]But this contradicts ( L geq S_1 > 1 ). Therefore, the assumption that ( S_k ) converges is false. Hence, ( S_k ) must diverge to infinity.Therefore, ( S_k to infty ) as ( k to infty ).Therefore, the terms ( x_k = frac{S_{k-1}}{S_{k-1} - 1} = 1 + frac{1}{S_{k-1} - 1} ). Since ( S_{k-1} to infty ), ( frac{1}{S_{k-1} - 1} to 0 ), so ( x_k to 1 + 0 = 1 ).Hence, the sequence ( x_n ) converges to 1.But let me check if this holds for the example I computed earlier with ( x_1 = 2 ):( x_1 = 2 )( x_2 = 2 / (2 - 1) = 2 )Wait, hold on, that's different from before. Wait, when I took ( x_1 = 2 ), computed ( x_2 = 2 / (2 - 1) = 2 ). Then, ( S_2 = x_1 + x_2 = 4 ), but also ( x_1 x_2 = 4 ). Then, ( x_3 = S_2 / (S_2 - 1) = 4 / 3 ≈ 1.333 ). Then, ( S_3 = S_2 + x_3 = 4 + 4/3 = 16/3 ≈ 5.333 ), which is equal to ( x_1 x_2 x_3 = 2 * 2 * 4/3 = 16/3 ). Then, ( x_4 = S_3 / (S_3 - 1) = (16/3) / (13/3) = 16/13 ≈ 1.230 ). Then, ( S_4 = S_3 + x_4 ≈ 16/3 + 16/13 ≈ (208 + 48)/39 ≈ 256/39 ≈ 6.564 ). Then, ( x_5 = S_4 / (S_4 - 1) ≈ 256/39 / (217/39) = 256/217 ≈ 1.179 ). Then, ( S_5 ≈ 256/39 + 256/217 ≈ ... ). Each subsequent term ( x_n ) is getting closer to 1. So even though ( S_k ) is increasing to infinity, the terms ( x_n ) are decreasing towards 1. Therefore, the sequence ( x_n ) is decreasing and bounded below by 1, so it should converge to 1.Wait, but in this case, the first few terms are 2, 2, 1.333, 1.230, 1.179, etc., approaching 1. So the sequence is decreasing after the second term.Wait, but starting from ( x_1 = 2 ), ( x_2 = 2 ), then ( x_3 = 4/3 ≈ 1.333 ), which is less than ( x_2 ). Then ( x_4 ≈ 1.230 ), which is less than ( x_3 ), etc. So the sequence is decreasing starting from ( x_2 ).But in the problem statement, ( x_1 > 1 ). So depending on ( x_1 ), maybe the behavior is similar.Let me check with a different ( x_1 ), say ( x_1 = 3 ).Then, ( x_2 = 3 / (3 - 1) = 1.5 )Then, ( S_2 = 3 + 1.5 = 4.5 = 3 * 1.5 = 4.5 ). Then, ( x_3 = 4.5 / (4.5 - 1) = 4.5 / 3.5 ≈ 1.2857 )Then, ( S_3 = 4.5 + 1.2857 ≈ 5.7857 ), which equals ( 3 * 1.5 * 1.2857 ≈ 5.7857 ). Then, ( x_4 = 5.7857 / (5.7857 - 1) ≈ 5.7857 / 4.7857 ≈ 1.209 )Similarly, ( x_4 ≈ 1.209 ), then ( x_5 ≈ S_4 / (S_4 - 1) approx (5.7857 + 1.209) / (5.7857 + 1.209 - 1) ≈ 6.9947 / 6.9947 ≈ Wait, no. Wait, S_4 = S_3 + x_4 ≈ 5.7857 + 1.209 ≈ 6.9947, so x_5 = 6.9947 / (6.9947 - 1) ≈ 6.9947 / 5.9947 ≈ 1.166.Then, x_5 ≈ 1.166, and so on. So again, the terms are decreasing towards 1. So in this case, the sequence starts with x_1=3, x_2=1.5, x_3≈1.2857, x_4≈1.209, x_5≈1.166, etc., approaching 1. So regardless of x_1 >1, x_2 = x_1/(x_1 -1). If x_1 is larger, x_2 is closer to 1. For example, if x_1=100, then x_2=100/99≈1.0101, then x_3≈ (100 + 100/99)/ ( (100 + 100/99) -1 )≈ (100 +1.0101)/ (100 +1.0101 -1 )≈101.0101 /100.0101≈1.0100, so x_3≈1.01, and so on, getting closer to 1 each time.Therefore, the sequence ( x_n ) is decreasing (after the first term if x_2 < x_1) and bounded below by 1. Therefore, by the Monotone Convergence Theorem, it must converge. Since we already showed that if the limit exists, it must be 1. Therefore, the limit is 1.But let's formalize this argument.First, show that ( x_n ) is decreasing for ( n ge 2 ).Given that ( x_{k} = frac{S_{k-1}}{S_{k-1} - 1} ). Since ( S_{k-1} ) is increasing, ( S_{k-1} - 1 ) is increasing as well. Let's see if ( x_{k} ) is decreasing.Compare ( x_{k+1} ) and ( x_k ):( x_{k+1} = frac{S_k}{S_k - 1} )( x_k = frac{S_{k-1}}{S_{k-1} - 1} )We need to show ( x_{k+1} < x_k ), i.e.,( frac{S_k}{S_k - 1} < frac{S_{k-1}}{S_{k-1} - 1} )Cross-multiplying (since all denominators and numerators are positive):( S_k (S_{k-1} - 1) < S_{k-1} (S_k - 1) )Expanding both sides:Left: ( S_k S_{k-1} - S_k )Right: ( S_{k-1} S_k - S_{k-1} )Subtract left from right:( (S_{k-1} S_k - S_{k-1}) - (S_k S_{k-1} - S_k) = -S_{k-1} + S_k )Which is ( S_k - S_{k-1} ). Since ( S_k > S_{k-1} ), this is positive. Therefore, the right side minus left side is positive, which implies the inequality holds. Therefore, ( x_{k+1} < x_k ).Therefore, starting from ( k=2 ), the sequence ( x_k ) is strictly decreasing. Since ( x_k > 1 ) for all ( k ) (since ( S_{k-1} >1 ), so ( x_k = S_{k-1}/(S_{k-1} - 1) = 1 + 1/(S_{k-1} -1 ) >1 )), the sequence is bounded below by 1. Hence, by the Monotone Convergence Theorem, ( x_k ) converges.Let ( lim_{k to infty} x_k = L ). We need to find L.But since ( x_k = frac{S_{k-1}}{S_{k-1} - 1} ), and ( S_{k-1} to infty ), then:( lim_{k to infty} x_k = lim_{S to infty} frac{S}{S - 1} = 1 ).Therefore, the limit is 1.Alternatively, since ( x_k = 1 + frac{1}{S_{k-1} - 1} ), and ( S_{k-1} to infty ), ( frac{1}{S_{k-1} - 1} to 0 ), so ( x_k to 1 + 0 = 1 ).Hence, the sequence ( x_n ) converges to 1.Final Answerboxed{1}</think>Given the sequence ( (x_n)_{n ge 1} ) with ( x_1 > 1 ) and satisfying the equation:[x_1 + x_2 + cdots + x_{n+1} = x_1 x_2 cdots x_{n+1} quad forall n in mathbb{N},]we need to show that this sequence is convergent and find its limit.First, we observe that for each ( n ), the sum of the first ( n+1 ) terms equals the product of those terms. Starting with ( n = 1 ), we have:[x_1 + x_2 = x_1 x_2 implies x_2 = frac{x_1}{x_1 - 1}.]For ( n = 2 ):[x_1 + x_2 + x_3 = x_1 x_2 x_3 implies x_3 = frac{x_1 x_2}{x_1 x_2 - 1}.]By induction, we find that each subsequent term ( x_{k+1} ) can be expressed as:[x_{k+1} = frac{S_k}{S_k - 1},]where ( S_k = x_1 + x_2 + cdots + x_k = x_1 x_2 cdots x_k ).We analyze the behavior of ( S_k ). The recursion relation for ( S_k ) is:[S_{k} = frac{S_{k-1}^2}{S_{k-1} - 1}.]We show that ( S_k ) is strictly increasing and tends to infinity. By induction, ( S_k ) is increasing. Assuming ( S_k ) converges to a finite limit ( L ) leads to a contradiction, implying ( S_k to infty ).Given ( S_k to infty ), the terms ( x_k ) can be written as:[x_k = frac{S_{k-1}}{S_{k-1} - 1} = 1 + frac{1}{S_{k-1} - 1}.]As ( S_{k-1} to infty ), the term ( frac{1}{S_{k-1} - 1} to 0 ), so ( x_k to 1 ).Additionally, we show that ( x_n ) is decreasing for ( n ge 2 ) and bounded below by 1, hence convergent by the Monotone Convergence Theorem. The limit is found to be 1.Thus, the sequence ( (x_n) ) converges to:[boxed{1}]]

🔑:1. Initial Setup and Definitions: We are given a sequence ( (x_n)_{n ge 1} ) with ( x_1 > 1 ) and the recurrence relation: [ x_1 + x_2 + cdots + x_{n+1} = x_1 x_2 cdots x_{n+1}, quad forall n in mathbb{N}. ] We need to show that this sequence is convergent and find its limit.2. Proving ( x_n > 1 ) for all ( n ): We start by assuming ( x_1, x_2, ldots, x_n > 1 ). For ( n = 1 ), this is given as ( x_1 > 1 ). Suppose ( x_1 + x_2 + cdots + x_n > x_1 x_2 cdots x_n - 1 ). We need to show that ( x_{n+1} > 1 ).3. Using Sum and Product Notations: Let (sigma = x_1 + x_2 + cdots + x_n) and (pi = x_1 x_2 cdots x_n). The recurrence relation gives: [ x_{n+1} = frac{sigma}{pi - 1}. ] Since ( x_1, x_2, ldots, x_n > 1 ), we have (pi > sigma). Therefore, (pi - 1 > sigma - 1), and: [ x_{n+1} = frac{sigma}{pi - 1} > frac{sigma}{sigma - 1} > 1. ]4. Establishing the Recurrence Relation for ( t_n ): Define ( t_n = frac{1}{x_n} ). Then ( t_n in (0, 1) ) for all ( n ge 1 ). From the recurrence relation, we have: [ t_{n+1} = t_n^2 - t_n + 1. ] We need to show that ( t_n ) is strictly increasing and converges to a limit ( l in (0, 1] ).5. Analyzing the Limit of ( t_n ): Since ( t_n in (0, 1) ) and ( t_{n+1} = t_n^2 - t_n + 1 ), we analyze the fixed points of the function ( f(t) = t^2 - t + 1 ). Solving ( l = l^2 - l + 1 ), we get: [ l^2 - 2l + 1 = 0 implies (l - 1)^2 = 0 implies l = 1. ] Therefore, ( t_n to 1 ) as ( n to infty ).6. Convergence of ( x_n ): Since ( t_n = frac{1}{x_n} ) and ( t_n to 1 ), it follows that ( x_n to 1 ) as ( n to infty ).The final answer is ( boxed{1} ).

🧠:[<think>Okay, let's tackle this problem where we need to find all triples of prime numbers p, q, r such that the absolute differences between each pair, |q - p|, |r - q|, and |r - p|, are also primes. Hmm, interesting. So, we need three primes where each pair's difference is a prime as well. Let me try to break this down step by step.First off, primes are numbers greater than 1 that have no divisors other than 1 and themselves. The primes start with 2, 3, 5, 7, 11, 13, and so on. Now, the differences between primes can sometimes also be primes. For example, 5 - 2 = 3, which is prime. But we need three primes where all three differences are primes. That seems more restrictive.Let me start by considering the smallest primes, since working with smaller numbers might make the problem more manageable. Let's note that 2 is the only even prime, and all other primes are odd. The difference between two odd primes will be even (since odd - odd = even), and the only even prime is 2. Therefore, if we have two odd primes, their difference must be 2 for it to be prime. This gives us twin primes (primes that differ by 2). So, maybe twin primes could play a role here.But in our problem, we need three primes p, q, r such that |q - p|, |r - q|, and |r - p| are all primes. Let's try to structure this.Suppose we start with p = 2, the smallest prime. Then, let's think about q. If p = 2, then |q - 2| must be prime. Let's consider possible q values. If q is an odd prime, then |q - 2| would be an odd number minus 2, which is odd. So, to be prime, the difference has to be an odd prime. Let's try q = 5. Then |5 - 2| = 3, which is prime. Good. Then we need to find r such that |r - 5| and |r - 2| are both primes. Let's pick r = 7. Then |7 - 5| = 2 (prime) and |7 - 2| = 5 (prime). So, here we have p = 2, q = 5, r = 7. Let's check all differences: 5-2=3, 7-5=2, 7-2=5. All primes. So that's one triple: (2, 5, 7).Wait, what about ordering? The problem says triples of primes p, q, r. So, are we supposed to consider permutations, or is there a specific order implied? The problem doesn't specify order, just "all triples". So maybe there are other permutations. But in the example above, p=2, q=5, r=7. If we reorder them, say p=5, q=2, r=7, then |2 - 5| = 3, |7 - 2| = 5, |7 - 5| = 2. Those are still primes, but the problem might consider ordered triples. Wait, the problem says "all triples of prime numbers p, q, r", so probably ordered triples. But maybe not. Wait, but in mathematics, a triple is usually ordered unless specified otherwise. However, the problem might consider unordered triples. Hmm, this is a bit ambiguous. Let me check the original problem again.The original problem states: "Find all triples of prime numbers p, q, r such that the numbers |q-p|, |r-q|, |r-p| are also prime." It doesn't specify if the order matters. But since it's talking about triples (p, q, r), by default, triples are ordered. So, (2,5,7) is different from (5,2,7), but in both cases, the differences are primes. Wait, but in the second case, |q-p| would be |2 - 5| = 3, which is prime, |r - q| = |7 - 2| = 5, which is prime, |r - p| = |7 - 5| = 2, which is prime. So, actually, both ordered triples (2,5,7) and (5,2,7) would satisfy the conditions. So maybe there are multiple ordered triples. However, perhaps the problem is looking for distinct primes, but even then, if the order matters, there could be multiple solutions. Alternatively, if the problem wants unordered sets, then (2,5,7) is the same as (5,2,7) etc. However, the problem says "triples", which are ordered. So perhaps we need to list all ordered triples where the differences are primes. But maybe the problem allows for p, q, r to be in any order, but the answer might list them in some order. Hmm. Let me proceed carefully.First, let's stick to the ordered triples. So, perhaps (2,5,7) is one, (5,2,7) is another, (2,7,5) might not work? Let me check. Let's take p=2, q=7, r=5. Then |7 - 2| = 5 (prime), |5 - 7| = 2 (prime), |5 - 2| = 3 (prime). So yes, that also works. So, permutations of (2,5,7) would work. So perhaps all permutations of these three primes would satisfy the condition. So, since there are 6 permutations, each of the 6 ordered triples would work. But maybe the problem is considering unordered triples, so just {2,5,7}. Hmm. The problem statement says "triples", but in math problems like this, sometimes they just want the set of primes, not considering order. But the problem says "triples of prime numbers p, q, r", so I think it's ordered. Therefore, the answer would include all permutations of 2,5,7. But let me check if there are other triples.Wait, maybe there are other triples. Let's check. Suppose we take p=2, q=3. Then |3 - 2| = 1, which is not prime. So that difference is invalid. So p=2, q=3 can't be part of such a triple. How about p=2, q=7. Then |7 - 2| = 5, which is prime. Then r must satisfy |r - 7| and |r - 2| are primes. Let's say r=5. Then |5 - 7| = 2, prime; |5 - 2| = 3, prime. So yes, (2,7,5) works. So that's similar to the previous case.Alternatively, take p=3, q=5. Then |5 - 3| = 2, prime. Then we need r such that |r - 5| and |r - 3| are primes. Let's try r=2. Then |2 - 5| = 3 (prime), |2 - 3| = 1 (not prime). So that doesn't work. How about r=7? |7 - 5| = 2 (prime), |7 - 3| = 4 (not prime). So that's invalid. How about r=3? Then |3 - 5| = 2 (prime), |3 - 3| = 0 (not prime). Not good. How about r=11? |11 - 5| = 6 (not prime), |11 - 3| = 8 (not prime). No. Hmm. So maybe starting with p=3 is problematic.What if we start with p=5? Let's take p=5, q=7. Then |7 - 5| = 2 (prime). Now, r needs to satisfy |r -7| and |r -5| are primes. Let's take r=2. Then |2 -7|=5 (prime), |2 -5|=3 (prime). So (5,7,2) is a valid triple. Similarly, (5,2,7) would also work as we saw before. So again, permutations of 2,5,7.Wait, let's check another starting point. Let's see if there are triples that don't include 2. Suppose all primes are odd. Then all differences would be even numbers (since odd - odd = even). The only even prime is 2. Therefore, all differences must be 2. So, if we have three primes p, q, r, all odd, then |q - p|, |r - q|, |r - p| must all be 2. But that would require p, q, r to form an arithmetic progression with difference 2. However, if |q - p| = 2 and |r - q| = 2, then r - q = ±2 and q - p = ±2. If they are both +2, then r - p = 4, which is not prime. If one is +2 and the other is -2, then p, q, r would not be in order. Let's see:Suppose p, q, r are odd primes. Then:Case 1: q - p = 2, r - q = 2. Then r - p = 4. But 4 is not prime, so invalid.Case 2: q - p = 2, r - q = -2. Then r = q - 2 = p + 2 - 2 = p. But then r = p, but primes must be distinct? Wait, the problem doesn't specify that the primes have to be distinct. Wait, but if p, q, r are primes, but they can be the same? Wait, primes are usually considered distinct in such problems unless repetition is allowed. Hmm, the problem says "triples of prime numbers p, q, r". So repetition is allowed? For example, could we have p = q = r = 2? Then the differences would be 0, 0, 0, which are not primes. So that's invalid.Alternatively, if two primes are the same. For example, p = q = 2, r = 3. Then |2 - 2| = 0 (not prime), |3 - 2| = 1 (not prime), |3 - 2| = 1. So invalid.So perhaps the primes must be distinct. The problem doesn't specify, but since 0 and 1 are not primes, the differences would have to be at least 2. So, maybe the primes in the triple must be distinct. Let's assume that.Therefore, if all three primes are odd, then their pairwise differences are even numbers greater than or equal to 2. The only even prime is 2. Therefore, the differences must all be 2. But as above, this leads to a contradiction unless the primes are in an arithmetic progression of difference 2, but then the difference between the first and last would be 4, which is not prime.Therefore, such a triple with all odd primes cannot exist. Hence, one of the primes must be 2. Since 2 is the only even prime, including 2 would result in some differences being odd and some even.Therefore, in any valid triple, exactly one of the primes is 2, and the other two are odd primes. Because if there were two even primes, they would both have to be 2, but then their difference would be 0, which isn't prime. So, only one 2, and the other two primes are odd.So, let's formalize this. Let me suppose that one of the primes is 2. Let's set p = 2. Then, q and r are odd primes. Then |q - 2|, |r - q|, and |r - 2| must all be primes.We need to find odd primes q and r such that:1. |q - 2| is prime.2. |r - q| is prime.3. |r - 2| is prime.Since q is an odd prime greater than 2 (since 2 is already taken as p), q is at least 3. Then |q - 2| is q - 2, which must be prime. So, q - 2 is prime. Similarly, |r - 2| must be prime, so r - 2 is prime (since r > 2, as it's an odd prime). Also, |r - q| must be prime. Since both q and r are odd primes, their difference is even (if they're both odd), unless one of them is 2. But since we already have p=2, and q and r are odd, their difference is even. So |r - q| must be 2, because that's the only even prime. Therefore, |r - q| = 2. Therefore, r and q are twin primes.So, putting this together:Given p = 2, then q must be a prime such that q - 2 is prime (so q is a prime that is 2 more than another prime), and r must be a prime such that r - 2 is prime (so r is a prime that is 2 more than another prime), and also |r - q| = 2. Therefore, q and r are twin primes. But q and r must also satisfy that q - 2 and r - 2 are primes.Wait, let's make this precise. Let's write down the conditions:1. q - 2 is prime.2. r - q = ±2 (since |r - q| = 2).3. r - 2 is prime.So, let's first consider r - q = 2. Then r = q + 2. Then:From condition 1: q - 2 is prime.From condition 3: (q + 2) - 2 = q is prime. But q is already prime. So that just tells us q is prime, which it is. But we already have q is prime, so this is redundant.Wait, so if r = q + 2, then r - 2 = q, which is prime. Therefore, as long as q is a prime such that q - 2 is also prime (so q is part of a pair of twin primes with the lower prime being q - 2), and r = q + 2 is also prime, then this satisfies the conditions. However, we need to make sure that in this case, |r - q| = 2, which is prime.Alternatively, if r - q = -2, then r = q - 2. Then:From condition 1: q - 2 is prime (so q is a prime such that q - 2 is prime).From condition 3: (q - 2) - 2 = q - 4 must be prime. So, in this case, q must be such that q - 2 and q - 4 are primes.Therefore, we have two cases:Case 1: r = q + 2.In this case, with p = 2, q is a prime where q - 2 is prime, and r = q + 2 is prime. But r = q + 2 is already implied by the twin prime condition. So, essentially, q must be a prime such that q - 2 is prime, and q + 2 is also prime. That is, q is the middle prime in a triplet of primes spaced by 2. However, such prime triplets (except for (3, 5, 7)) are rare. Wait, let's check.For example, starting with p = 2:If q = 5, then q - 2 = 3 (prime), and r = q + 2 = 7 (prime). So, this gives us the triple (2, 5, 7), which we already saw.Another example: Let's take q = 7. Then q - 2 = 5 (prime), and r = 9, which is not prime. So that's invalid.q = 3: q - 2 = 1, which is not prime. So invalid. So the only q that works here is 5.Wait, but if q = 3, then q - 2 = 1, which isn't prime, so q = 3 doesn't work. So the only q in this case is 5, leading to r = 7. So that gives us (2, 5, 7).Case 2: r = q - 2.In this case, with p = 2, q must satisfy q - 2 is prime (from condition 1), and r = q - 2 must satisfy r - 2 = q - 4 is prime. So, q must be a prime such that q - 2 and q - 4 are both primes.Let's check possible q:q = 7: q - 2 = 5 (prime), q - 4 = 3 (prime). So r = 5. Then, checking the differences:|q - p| = |7 - 2| = 5 (prime),|r - q| = |5 - 7| = 2 (prime),|r - p| = |5 - 2| = 3 (prime).So, that works. So (2, 7, 5) is another ordered triple. Similarly, if we take q = 5, then q - 2 = 3 (prime), q - 4 = 1 (not prime). So that doesn't work. q = 11: q - 2 = 9 (not prime). q = 13: q - 2 = 11 (prime), q - 4 = 9 (not prime). So, only q =7 works here. Therefore, this gives us another triple: (2, 7, 5).So, combining both cases, when p=2, we get two ordered triples: (2,5,7) and (2,7,5). But wait, depending on how we assign q and r. Wait, in the first case, q=5, r=7. In the second case, q=7, r=5. So, these are two different ordered triples.But are there other possibilities? Let's see.Suppose p=2, q=5, r=7.Another case: p=2, q=7, r=5.But maybe if we set another prime as 2. For example, set q=2 instead of p=2. Let's try that.Set q=2. Then we need |2 - p|, |r - 2|, and |r - p| to be primes.So, p and r are primes (could be even or odd, but since q=2 is even, let's see):Since p is a prime, and |2 - p| must be prime. If p is an odd prime, then |2 - p| is |odd - even| = odd, so the difference is odd. Therefore, to be prime, it must be an odd prime (so not 2). So, |2 - p| is prime. Let's take p=5. Then |2 -5| =3, which is prime. Then, |r - 2| must be prime. Let's take r=7. Then |7 -2|=5 (prime). Then |7 -5|=2 (prime). So, that works. So here we have (5,2,7). Similarly, if we take p=7, |2 -7|=5 (prime), then r needs to satisfy |r -2| and |r -7| primes. Let's take r=5. Then |5 -2|=3 (prime), |5 -7|=2 (prime). So, (7,2,5) is another ordered triple.Alternatively, if we set r=2, let's see. Let's set r=2. Then, we need |q - p|, |2 - q|, and |2 - p| to be primes.So, p and q are primes (could be 2 or other primes). Let's see:First, |2 - p| must be prime. If p is an odd prime, then |2 - p| is an odd number, so must be an odd prime. If p=3, then |2 -3|=1 (not prime). If p=5, |2 -5|=3 (prime). If p=7, |2 -7|=5 (prime). Similarly, |2 - q| must be prime. So q must be a prime such that |2 - q| is prime. Let's take p=5 and q=7. Then |7 -5|=2 (prime), |2 -7|=5 (prime), |2 -5|=3 (prime). So, (5,7,2) is another ordered triple. Similarly, if p=7 and q=5, then |5 -7|=2 (prime), |2 -5|=3 (prime), |2 -7|=5 (prime). So, (7,5,2) is another ordered triple.Therefore, by permuting the positions of 2, 5, and 7, we can get all six ordered triples: (2,5,7), (2,7,5), (5,2,7), (5,7,2), (7,2,5), (7,5,2). Each of these permutations satisfies the conditions. So, if the problem is asking for ordered triples, then all six permutations are solutions. However, if the problem is asking for unordered sets, then the only solution is {2,5,7}. But given that the problem mentions "triples", which are ordered, I think the answer expects all ordered permutations. But let me check if there are other triples beyond permutations of 2,5,7.Let me check if there's another set of primes. Suppose we take p=2, q=3. Then |3 -2|=1, which is not prime. So invalid. If p=2, q=7, r=3. Then |7-2|=5, |3-7|=4, which is not prime. So invalid. If p=2, q=11. Then |11 -2|=9, which is not prime. So invalid. Similarly, p=2, q=13. |13-2|=11, which is prime. Then r must satisfy |r -13| and |r -2| are primes. Let's try r=11. |11 -13|=2 (prime), |11 -2|=9 (not prime). r=17. |17 -13|=4 (not prime). r=19. |19 -13|=6 (not prime). r=7. |7 -13|=6 (not prime). r=5. |5 -13|=8 (not prime). So, even if q=13 and p=2, there is no r that satisfies the conditions. Therefore, this approach doesn't yield a new triple.Alternatively, suppose we take p=2, q=5, r=3. Then |5 -2|=3, |3 -5|=2, |3 -2|=1 (not prime). So invalid. How about p=2, q=5, r=11. |5 -2|=3, |11 -5|=6 (not prime). Invalid. p=2, q=5, r=3. As above, invalid. p=2, q=5, r=7. Works.What about another triple with a different starting prime? For example, p=3. Let's try p=3, then |q -3|, |r - q|, |r -3| must all be primes. Let's try q=5. |5 -3|=2 (prime). Then |r -5| and |r -3| must be primes. Let's take r=7. |7 -5|=2 (prime), |7 -3|=4 (not prime). So invalid. Take r=2. |2 -5|=3 (prime), |2 -3|=1 (not prime). Take r=11. |11 -5|=6 (not prime). Take r=3. |3 -5|=2 (prime), |3 -3|=0 (not prime). Doesn't work. So p=3, q=5 doesn't work. How about q=7? |7 -3|=4 (not prime). So invalid. So p=3 doesn't seem to lead to a valid triple.Alternatively, p=5. Let's set p=5. Then |q -5| must be prime. Let's take q=7. |7 -5|=2 (prime). Then |r -7| and |r -5| must be primes. Take r=2. |2 -7|=5 (prime), |2 -5|=3 (prime). So, (5,7,2) works. Similarly, take r=3. |3 -7|=4 (not prime). So invalid. Take r=12. Not prime. So, only r=2 works here. Therefore, (5,7,2) is a valid triple, which is a permutation of our earlier solution.Alternatively, take p=5, q=2. |2 -5|=3 (prime). Then |r -2| and |r -5| must be primes. Take r=7. |7 -2|=5 (prime), |7 -5|=2 (prime). So (5,2,7) works. So again, permutation of 2,5,7.Another case, p=7. Let's set p=7. Then |q -7| must be prime. Take q=5. |5 -7|=2 (prime). Then |r -5| and |r -7| must be primes. Take r=2. |2 -5|=3 (prime), |2 -7|=5 (prime). So (7,5,2) is valid. Similarly, take q=2. |2 -7|=5 (prime). Then |r -2| and |r -7| must be primes. Take r=5. |5 -2|=3 (prime), |5 -7|=2 (prime). So, (7,2,5) is valid.So, in all cases, the only triples that work are permutations of (2,5,7). Therefore, all ordered triples where the primes are 2, 5, and 7 in some order. Since there are 3! = 6 permutations, there are six ordered triples. However, if the problem expects unordered triples, then only {2,5,7} is the solution.But let's verify once more if there's another set of primes. Let's try p=2, q=5, r=11. Then differences are |5-2|=3 (prime), |11-5|=6 (not prime), |11-2|=9 (not prime). Doesn't work. How about p=2, q=7, r=11. |7-2|=5, |11-7|=4 (not prime). Nope. p=2, q=11, r=13. |11-2|=9 (not prime). Invalid. How about p=2, q=3, r=5. |3-2|=1 (invalid). p=2, q=3, r=7. |3-2|=1 (invalid). p=2, q=3, r=5. Same.Alternatively, perhaps including the prime 3. Let me see. Let's take p=3, q=5, r=7. Then differences: |5-3|=2 (prime), |7-5|=2 (prime), |7-3|=4 (not prime). So invalid. How about p=3, q=7, r=5. |7-3|=4 (invalid). p=5, q=3, r=7. |3-5|=2 (prime), |7-3|=4 (invalid). So, 3 doesn't seem to help.Another thought: if one of the primes is 2, and the other two differ by 2 (twin primes), but also spaced such that their differences with 2 are primes. For example, if the other primes are 5 and 7, which are twin primes (difference 2), and 5 - 2 = 3 (prime), 7 - 2 = 5 (prime). So that works. What about another pair of twin primes? Let's take 11 and 13. Then 11 - 2 = 9 (not prime), so that doesn't work. 17 and 19: 17 - 2 =15 (not prime). So, no. The only twin primes where both are 2 more than a prime are 5 and 7. Because 5 - 2 = 3 (prime), 7 - 2 =5 (prime). Similarly, check 3 and 5: 3 -2=1 (not prime), so invalid. Hence, only 5 and 7 satisfy that when paired with 2.Therefore, the only triple possible is permutations of (2,5,7). Therefore, the answer should be all permutations of 2,5,7. Since the problem asks for triples of primes, and triples are ordered, the answer includes all six ordered permutations.But wait, let me check once more if there exists another triple with different primes. Let me think of primes like 2, 3, 5. Let's see. |3 -2|=1 (invalid). So that's out. 2,7,11. |7 -2|=5 (prime), |11 -7|=4 (invalid). 2, 11,13. |11-2|=9 (invalid). 2,5,3. |5 -2|=3, |3 -5|=2, |3 -2|=1 (invalid). 2,3,7. |3-2|=1, invalid. 2,5,11. |5-2|=3, |11-5|=6 (invalid). Nope. 2,7,5. Already covered. 3,5,7. Differences: 2,2,4 (invalid). 5,2,7. Covered. 7,5,2. Covered. Etc.Therefore, after checking various possibilities, it seems that the only possible triples are permutations of 2,5,7. Hence, the answer is all ordered triples where the primes are 2,5,7 in any order. Therefore, there are six such triples: (2,5,7), (2,7,5), (5,2,7), (5,7,2), (7,2,5), (7,5,2).However, sometimes in such problems, the order might not matter, and they just want the set {2,5,7}. But given the problem states "triples", which are ordered, and asks for "all triples", I think all six permutations are required. However, let me check the problem statement again: "Find all triples of prime numbers p, q, r such that the numbers |q-p|,|r-q|,|r-p| are also prime."It doesn't specify if the triples are ordered or unordered, but in mathematical terms, a triple is an ordered collection. However, in some contexts, "triple" can refer to a set. But given that the problem is about differences between pairs, which would be the same regardless of order, but since the differences are |q-p|, |r-q|, |r-p|, which depend on the order of p, q, r, the problem might indeed be expecting ordered triples. For example, in the triple (5,7,2), the differences would be |7-5|=2, |2-7|=5, |2-5|=3, which are all primes. So, even though the set {2,5,7} is the same, the ordered triples are different but all satisfy the conditions.Therefore, the conclusion is that all ordered triples consisting of the primes 2,5,7 in any order satisfy the given conditions, and there are no other triples. Hence, the answer is all permutations of (2,5,7).Final AnswerThe triples are all permutations of (2, 5, 7). Thus, the solutions are boxed{(2, 5, 7)}, boxed{(2, 7, 5)}, boxed{(5, 2, 7)}, boxed{(5, 7, 2)}, boxed{(7, 2, 5)}, and boxed{(7, 5, 2)}.</think>To solve the problem of finding all triples of prime numbers ( p, q, r ) such that the absolute differences ( |q - p| ), ( |r - q| ), and ( |r - p| ) are also primes, we start by considering the properties of prime numbers and their differences.1. Key Observations: - The only even prime number is 2. All other primes are odd. - The difference between two odd primes is even, and the only even prime is 2. Therefore, if two odd primes differ by 2, they are twin primes. - Including the prime number 2 in the triple is crucial because it allows for odd differences, which can be prime.2. Considering the Prime 2: - If one of the primes is 2, the other two primes must be odd. The differences involving 2 must be prime. - We need to find primes ( q ) and ( r ) such that ( |q - 2| ), ( |r - q| ), and ( |r - 2| ) are all primes.3. Testing Possible Triples: - Let ( p = 2 ). Then ( q ) and ( r ) must be odd primes such that ( |q - 2| ), ( |r - q| ), and ( |r - 2| ) are primes. - Trying ( q = 5 ) and ( r = 7 ): - ( |5 - 2| = 3 ) (prime) - ( |7 - 5| = 2 ) (prime) - ( |7 - 2| = 5 ) (prime) - This gives the triple ( (2, 5, 7) ).4. Checking Permutations: - By permuting the primes 2, 5, and 7, we check all ordered triples: - ( (2, 5, 7) ) - ( (2, 7, 5) ) - ( (5, 2, 7) ) - ( (5, 7, 2) ) - ( (7, 2, 5) ) - ( (7, 5, 2) ) - All these permutations satisfy the condition that the absolute differences are primes.5. Conclusion: - No other sets of primes satisfy the conditions because including other primes or different configurations results in non-prime differences.Thus, the solutions are all permutations of the primes 2, 5, and 7. The final answer is:[boxed{(2, 5, 7)}, boxed{(2, 7, 5)}, boxed{(5, 2, 7)}, boxed{(5, 7, 2)}, boxed{(7, 2, 5)}, boxed{(7, 5, 2)}],<think>Okay, so I need to find all triples of prime numbers p, q, r such that the absolute differences between each pair, |q - p|, |r - q|, and |r - p|, are also prime numbers. Hmm, let's start by understanding the problem. First, primes are numbers greater than 1 that have no divisors other than 1 and themselves. The smallest primes are 2, 3, 5, 7, 11, etc. Now, the differences between these primes also need to be primes. Since we're dealing with absolute differences, they must be positive primes. So, the differences could be 2, 3, 5, 7, etc. But here's a thought: the only even prime is 2. All other primes are odd. If we subtract two odd primes, the difference will be even (since odd - odd = even). The only even prime is 2. So, unless the two primes are 2 apart, their difference will be even and composite (except for 2). Therefore, in most cases, the differences between primes (other than 2) will be even numbers, which are not prime unless the difference is exactly 2. This makes me think that in order for |q - p|, |r - q|, and |r - p| to all be primes, most of these differences have to be 2. However, since all three differences are primes, perhaps they could include 2 and some other primes. Let's note that the differences can't be 1, because 1 is not prime. Let me start by considering small primes first. Let's try with the smallest primes. Let's see if there's a triple that includes 2, since 2 is the only even prime. If we include 2, then the other primes must be odd, so their differences from 2 will be odd - 2, which is odd. But 2 is even, so |q - p| could be 2 if the primes are 2 apart, or it could be another prime if the difference is odd. Wait, no. If one prime is 2 and the other is an odd prime, then the difference is |odd - 2|. If the odd prime is larger than 2, then the difference is odd - even = odd. So, for example, if we take p = 2 and q = 3, then |3 - 2| = 1, which is not prime. So that's bad. If p = 2 and q = 5, then |5 - 2| = 3, which is prime. So that works. Then moving on, if we have another prime r, let's say r = 7. Then |7 - 5| = 2, which is prime, and |7 - 2| = 5, which is prime. Wait, that seems to work. So here we have p = 2, q = 5, r = 7. Let's check:|q - p| = |5 - 2| = 3 (prime)|r - q| = |7 - 5| = 2 (prime)|r - p| = |7 - 2| = 5 (prime)So that works. Okay, so that's one triple: (2, 5, 7). Let me check if there are others. What if we start with p = 2, then q must be a prime such that |q - 2| is prime. Let's list primes q where q - 2 is prime. So q can be 5 (5 - 2 = 3), 7 (7 - 2 = 5), 13 (13 - 2 = 11), etc. Similarly, primes q where q is less than 2 (but primes are positive, so the only prime less than 2 is none, since 2 is the smallest prime). So q must be greater than 2, and q - 2 is prime. So q is 2 + a prime. So the next prime after 2 is 3, but 3 - 2 = 1, which isn't prime. Then 5 - 2 = 3, which is prime. So q can be 5, 7, 13, etc. So let's take p = 2, q = 5. Then we need a prime r such that |r - 5| is prime and |r - 2| is prime. Let's try r = 7. As before, that works. What about r = 3? |3 - 5| = 2, which is prime, and |3 - 2| = 1, which is not prime. So that fails. How about r = 11? |11 - 5| = 6, which is not prime. |11 - 2| = 9, which is not prime. So that's no good. Next, r = 7: works. What about r = 5? Then |5 - 5| = 0, which isn't prime. So same number, but primes are distinct here? Wait, the problem says triples of prime numbers p, q, r. So are they allowed to repeat? The problem says triples of primes, but it doesn't specify distinct. Hmm. Wait, but if two primes are the same, then their difference is 0, which isn't prime. So even if repetition is allowed, those cases would fail because of the 0. So perhaps the primes must be distinct. So we can assume p, q, r are distinct primes. So p = 2, q = 5, r = 7 works. Let's check another q. Let's take p = 2, q = 7 (since 7 - 2 = 5, which is prime). Then we need to find r such that |r - 7| is prime and |r - 2| is prime. Let's try r = 11. |11 - 7| = 4 (not prime), |11 - 2| = 9 (not prime). r = 3: |3 - 7| = 4 (not prime), |3 - 2| = 1 (not prime). r = 5: |5 - 7| = 2 (prime), |5 - 2| = 3 (prime). So here, p = 2, q = 7, r = 5. But since we are talking about triples, the order might matter? Wait, the problem doesn't specify order. So if we have p, q, r as primes, but it's a triple, so ordered or unordered? The problem says "all triples", which in mathematics can sometimes mean ordered triples, but since the differences are taken between pairs regardless of order, maybe it's unordered. But in the example I had earlier, (2,5,7), arranging them in order would be 2,5,7. If I take (2,7,5), then the differences would be |7-2|=5, |5-7|=2, |5-2|=3, which are the same as before. So perhaps the triples are considered unordered. So (2,5,7) and (2,7,5) would be the same triple. Therefore, maybe all such triples are sets where the primes are arranged in order. So maybe we can fix the order as increasing, so p < q < r, to avoid repetition. So perhaps the triples are ordered such that p < q < r. Then the differences would be q - p, r - q, and r - p. Since all differences are positive, no absolute value needed. So in that case, we need three primes p < q < r such that q - p, r - q, and r - p are all primes. So in my previous example, p = 2, q = 5, r = 7. Then q - p = 3, r - q = 2, and r - p = 5. All primes. So that works. Are there others?Let's see. Let's try p = 2. Then q needs to be such that q - 2 is prime. As before, possible qs are 5, 7, 13, etc. Let's try q = 5. Then r needs to satisfy r - 5 is prime and r - 2 is prime. So r must be a prime such that r - 5 and r - 2 are primes. Let's see, r = 7: 7 - 5 = 2 (prime), 7 - 2 = 5 (prime). Good. r = 11: 11 - 5 = 6 (not prime), 11 - 2 = 9 (not prime). r = 13: 13 - 5 = 8 (not prime), 13 - 2 = 11 (prime). But 8 is not prime. r = 17: 17 - 5 = 12 (not prime). Hmm. So seems only r = 7 works here. What if p = 2, q = 7? Then r must satisfy r - 7 and r - 2 are primes. Let's try r = 11: 11 - 7 = 4 (not prime), r = 13: 13 - 7 = 6 (not prime), r = 5: 5 - 7 = -2, absolute value 2 (prime), and 5 - 2 = 3 (prime). So that would give the same triple as before but in reverse. But since we are considering p < q < r, then p = 2, q = 5, r =7. So even if we take p =2, q =7, r would have to be greater than 7, but then r -7 needs to be prime. Let's check r = 12, but 12 isn't prime. r =11: 11-7=4, not prime. r=13: 13-7=6, not prime. So that doesn't work. So maybe q =7 doesn't lead to any new triples when p=2.Alternatively, maybe try a different starting prime. Let's take p=3. Then q must satisfy q -3 is prime. So possible q's are primes such that q = 3 + prime. Let's see, primes starting from 2: 3 + 2 =5 (prime), 3 +3=6 (not prime), 3 +5=8 (not prime). So q=5 is the only one. Then r must satisfy r -5 is prime and r -3 is prime. So r needs to be a prime where r -5 and r -3 are primes. Let's try r=7: 7 -5=2 (prime), 7 -3=4 (not prime). So no. r=11: 11-5=6 (not prime). r=8: not prime. r=10: not prime. r=13: 13-5=8 (not prime). r=5: same as q, so difference 0. Doesn't work. So p=3 doesn't seem to lead to a valid triple.How about p=5? Then q needs to be 5 + prime. So possible q=5+2=7, which is prime. Then r needs to be such that r -7 and r -5 are primes. Let's try r=7: same as q, difference 0. r=11: 11-7=4 (not prime). r=13:13-7=6 (not prime). r=12: not prime. r=17:17-7=10 (not prime). So no luck here. Alternatively, q=5 +3=8 (not prime), so q must be 7. So p=5, q=7. Then r? 11: 11-7=4, not prime. 7-5=2, but r needs to be greater. So maybe p=5 is not helpful.Wait, maybe trying p=2 is the only way since 2 is the only even prime, and the differences involving 2 can be odd primes. Let me think. If we have p=2, then the differences q-p and r-p are odd primes. Then between q and r, the difference r - q must also be a prime. Since both q and r are odd primes, their difference is even, unless it's 2. Wait, odd - odd = even, so |r - q| must be 2, the only even prime, or another prime? But even numbers can't be primes unless they are 2. Therefore, |r - q| must be 2. Therefore, in the triple (p, q, r), if p=2, then q and r must be twin primes (i.e., primes that differ by 2). But also, the differences from 2 must be primes. Let me check this.If p=2, then q must be such that q - 2 is prime, and r must be such that r - q is 2 (since it has to be even and prime, so only 2) and r - 2 must be prime. So if r - q = 2, then r = q + 2. Then r - 2 = (q + 2) - 2 = q. So q must be prime. But r - 2 = q, which is prime. Therefore, if we have p=2, q, r=q+2, then:- |q - p| = q - 2 (must be prime)- |r - q| = 2 (prime)- |r - p| = r - 2 = q (prime)Therefore, the conditions reduce to: q must be a prime such that q - 2 is also prime. Then r = q + 2 must also be prime. So q must be a prime such that q - 2 and q + 2 are primes. These are called "prime triplets" of the form (q - 2, q, q + 2). The only prime triplet where all three are prime is (3, 5, 7). Because 3, 5, 7 are all primes. Let's check:- 5 - 2 = 3 (prime)- 7 - 5 = 2 (prime)- 7 - 2 = 5 (prime)So that works. Are there any other prime triplets? For example, starting at 5: 5, 7, 9. But 9 isn't prime. Starting at 7: 7, 9, 11. 9 isn't prime. Starting at 11: 11, 13, 15. 15 isn't prime. So the only triplet where all three are primes is (3, 5, 7). But in our case, p=2, so we have (2, 5, 7). Wait, why isn't (2, 3, 5) considered? Let's check:If p=2, q=3, r=5:- |3 - 2| = 1 (not prime)- |5 - 3| = 2 (prime)- |5 - 2| = 3 (prime)So since the first difference is 1, which is not prime, this doesn't work. Therefore, q=3 is invalid here. So only when q=5, because 5 - 2 = 3 (prime), and then r=7, which is 7 -5=2 (prime), and 7 -2=5 (prime). So that's the only triplet of primes where p=2 and q, r are part of a prime triplet.Therefore, the only such triple is (2, 5, 7). Wait, but let me check if there are other possibilities where p isn't 2. Suppose all three primes are odd. Then their differences would be even numbers (since odd - odd = even). So the differences must be 2, the only even prime. So in this case, if all three primes are odd, then the differences between each pair must be 2. But three primes can't all be 2 apart from each other unless they form a sequence like 3, 5, 7. Let's see:Take primes 3, 5, 7. Then:|5 - 3| = 2 (prime)|7 - 5| = 2 (prime)|7 - 3| = 4 (not prime)So the last difference is 4, which isn't prime. Therefore, this doesn't work. So even though 3,5,7 are primes with differences 2, the overall difference of 4 isn't prime. So this triple fails. Alternatively, if we have three primes where each adjacent pair differs by 2 (twin primes), but the overall difference is 4. So that seems not helpful. Alternatively, if we have three primes such that each pair differs by 2. But that's impossible unless they are 3,5,7. However, as shown, the difference between 3 and 7 is 4. So, no.Alternatively, maybe two primes differ by 2 and the third prime differs from one of them by another prime. Let's see. For example, take p=3, q=5, r=7. As before, differences: 2, 2, 4. 4 is not prime. What if p=3, q=7, then r=?|q - p| = 4 (not prime). So that fails immediately. Alternatively, p=5, q=7, r=11. Differences:|7 -5| = 2 (prime)|11 -7| =4 (not prime)|11 -5| =6 (not prime)Nope. Alternatively, p=5, q=11, r=13:|11 -5|=6 (not prime), so already fails.Alternatively, p=7, q=11, r=13:|11 -7|=4 (not prime)So that's not working. Alternatively, p=3, q=5, r=11:|5 -3|=2 (prime)|11 -5|=6 (not prime)|11 -3|=8 (not prime)Not good. So, if all three primes are odd, the differences between them will be even numbers. The only even prime is 2, so unless the differences are 2, they won't be prime. However, in such a case, you can have two primes differing by 2, but the third prime would have to differ from both by 2, which would require the three primes to form an arithmetic sequence of three primes with common difference 2. The only such triplet is 3,5,7, but as we saw, the difference between 3 and7 is4, which is not prime. Therefore, such a triplet doesn't work. Thus, the only possible triples must include the prime 2. Because if 2 is included, then the differences between 2 and the other primes can be odd primes, and the difference between the other two primes must be 2 (as it's even and only 2 is prime). So, as we saw earlier, starting with 2, the next prime q must be 2 + a prime. Let's call that prime d1, so q = 2 + d1, where d1 is prime. Then, the third prime r must be q + d2, where d2 is prime. But since r - q must be prime and even (because q is odd, r is odd, their difference is even), so d2 must be 2. Therefore, r = q + 2. Then, r - 2 must also be prime. But r - 2 = (q + 2) - 2 = q. So q must be prime. Which it is, by definition. Wait, but q is already prime. So if we have p=2, q=2 + d1 (prime), and r = q +2, then the conditions are:- q - p = d1 (prime)- r - q = 2 (prime)- r - p = q + 2 - 2 = q (prime)Therefore, all conditions are satisfied if q is a prime such that q - 2 is also prime, and r = q + 2 is prime. So q must be a prime such that both q - 2 and q + 2 are primes. These are called "isolated primes" or part of a prime triplet. The only such prime is 5, since 5 - 2 = 3 and 5 + 2 =7, both primes. Another example, take q=3: q -2=1 (not prime), so invalid. q=7: q -2=5 (prime), q +2=9 (not prime). So 7 fails. q=11: q -2=9 (not prime). So invalid. Therefore, the only q that works is 5. Hence, the only triple is p=2, q=5, r=7. Therefore, the only such triple is (2, 5, 7). Let me check if there could be another configuration where p is not the smallest prime, but another arrangement. For example, suppose we have p=2, r=5, and q=7. Then:|q - p| = |7 -2|=5 (prime)|r - q|=|5 -7|=2 (prime)|r - p|=|5 -2|=3 (prime)So this also works. But in terms of ordered triples, if we consider unordered sets, {2,5,7} is the same regardless of order. So depending on whether the problem requires ordered triples or unordered sets. The problem states "all triples of prime numbers p, q, r", so perhaps they are ordered triples. But in that case, we need to list all permutations that satisfy the condition. However, since the differences are absolute values, the order might not matter. Let me check:If we take p=5, q=2, r=7:|2 -5|=3 (prime)|7 -2|=5 (prime)|7 -5|=2 (prime)So that also works. So if we consider ordered triples, all permutations of 2,5,7 would work. However, in the problem statement, it's not specified if order matters. In mathematics, when asked for "triples", unless specified, sometimes they are considered ordered. However, given the problem is about differences which are absolute, the order of p, q, r might not affect the result. For example, in the triple (5,2,7), the differences would still be |2-5|=3, |7-2|=5, |7-5|=2. So regardless of the order, the set {2,5,7} would produce the required prime differences. Therefore, depending on the interpretation, the answer could be all permutations of (2,5,7), or just the set {2,5,7}. However, in most mathematical problems, unless specified otherwise, "triples" can be considered ordered. But since the differences are absolute and the problem doesn't specify order, it's possible that the answer is just the single unique set {2,5,7}. To confirm, let's see if there's another set. Suppose someone claims (2,3,5):|3-2|=1 (not prime)|5-3|=2 (prime)|5-2|=3 (prime)Since 1 is not prime, this fails. Another candidate: (2,7,11):|7-2|=5 (prime)|11-7|=4 (not prime)|11-2|=9 (not prime)Nope. Another candidate: (2,5, 11):|5-2|=3 (prime)|11-5|=6 (not prime)|11-2|=9 (not prime)No good. (2,5, 3): same as (2,3,5). (2, 5, 7): works. (2,7,5): same as (2,5,7). (5,2,7): same differences. So regardless of order, the only distinct set is {2,5,7}. Alternatively, if ordered triples are required, then all permutations where the differences are computed in order. For example, if the problem had said "p, q, r in that order such that |q - p|, |r - q|, and |r - p| are primes", then the order would matter. But given that they mention the differences |q - p|, |r - q|, |r - p|, which are three differences regardless of order, perhaps the triples are unordered. But let's check if there's another set with different primes. For example, could we have a triple like (3,7,11):|7-3|=4 (not prime)So no. (3,5,11):|5-3|=2 (prime)|11-5|=6 (not prime)|11-3|=8 (not prime)No. (5,7,11):|7-5|=2 (prime)|11-7|=4 (not prime)|11-5|=6 (not prime)No. (7,11,13):|11-7|=4 (not prime)Already fails. How about (2,5,7):As established, works.(2,5,7) is the only one. Wait, let me think again. Suppose we take p=2, q=5, r=7. What if we take another prime triplet? Let's suppose p=2, q= next prime, but not 5. For example, p=2, q=13 (since 13-2=11, which is prime). Then r would have to be q + 2 =15, but 15 isn't prime. So that fails. Similarly, q=11 (11-2=9, not prime). q=7 (7-2=5, prime). Then r=9 (not prime). So r=7 +2=9 invalid. So the only q where q-2 and q+2 are primes is q=5. Therefore, the only possible r is 7. Therefore, the only triple is (2,5,7). Is there a triple without 2? Let's check. Suppose all primes are odd. Then all differences are even, so must be 2. So we need three primes where each pair differs by 2. But as discussed earlier, the only three primes in an arithmetic progression with difference 2 are 3,5,7. However, the difference between 3 and7 is4, which is not prime, so that fails. Alternatively, if we have three primes where two differ by 2 and the third differs by another prime. For example, p=3, q=5, r=7. Differences: |5-3|=2 (prime), |7-5|=2 (prime), |7-3|=4 (not prime). So no. Alternatively, p=3, q=7, r= something. |7 -3|=4 (not prime). So already fails. Alternatively, p=5, q=7, r= something. |7 -5|=2 (prime). Let's say r=11. |11 -7|=4 (not prime). |11 -5|=6 (not prime). Doesn't work. Thus, no triples without 2. Therefore, the only possible triple is {2,5,7}. To confirm, let me check all possibilities once more. Suppose someone found another triple. Let's say (2,3,7). Then:|3-2|=1 (not prime). So nope. (2,7,11):|7-2|=5 (prime)|11-7|=4 (not prime). So nope. (2,5,11):|5-2|=3 (prime)|11-5|=6 (not prime). So no. (2,5,13):|5-2|=3 (prime)|13-5|=8 (not prime). No. (2,5,9): 9 not prime. (2,3,5): as before, fails. So, yeah, (2,5,7) is the only one. Therefore, the answer is the triple (2,5,7). Depending on the problem's requirement, if order matters, all permutations, but likely as an unordered set, so {2,5,7}. But in the problem statement, they might expect the ordered triple in ascending order, so (2,5,7). Final AnswerThe only such triple is boxed{(2, 5, 7)}.</think>To find all triples of prime numbers ( p, q, r ) such that the numbers ( |q - p|, |r - q|, |r - p| ) are also prime, we start by considering the properties of prime numbers and their differences. 1. Key Observations: - The only even prime number is 2. All other primes are odd. - The difference between two odd primes is even, and the only even prime is 2. Therefore, if the difference between two primes is prime, it must be 2 or another prime. - Including the prime number 2 is crucial because it allows for odd differences, which can be prime.2. Considering the Prime 2: - If ( p = 2 ), we need primes ( q ) and ( r ) such that ( |q - 2| ), ( |r - q| ), and ( |r - 2| ) are all primes. - Let ( q ) be a prime such that ( q - 2 ) is also prime. The primes ( q ) that satisfy this are 5, 7, 13, etc. However, we also need ( r ) such that ( r - q ) and ( r - 2 ) are primes.3. Checking Specific Cases: - For ( q = 5 ), we need ( r ) such that ( r - 5 ) and ( r - 2 ) are primes. Trying ( r = 7 ): - ( |5 - 2| = 3 ) (prime) - ( |7 - 5| = 2 ) (prime) - ( |7 - 2| = 5 ) (prime) - This works, so the triple ( (2, 5, 7) ) is valid.4. Considering Other Primes: - If all primes are odd, their differences would be even and must be 2. However, three primes differing by 2 would form a sequence like 3, 5, 7, but the difference between 3 and 7 is 4 (not prime), so this fails. - Trying other combinations of primes without including 2 results in non-prime differences.5. Conclusion: - The only valid triple that satisfies all conditions is ( (2, 5, 7) ). This is because the differences ( 3, 2, ) and ( 5 ) are all primes, and no other combinations of primes meet the required conditions.[boxed{(2, 5, 7)}]]

🔑:Answer: 2,5,7.Solution. We can assume that p2, then all numbers p, q, r are odd, and their differences |q-p|,|r-q|,|r-p| are even. Then |r-p|=r-p=r-q+q-r geq 2+2=4 cannot be a prime number, which contradicts the problem's condition. Therefore, p=2.Then q and r are odd prime numbers, and their difference r-q is an even prime number, so r-q=2, which means r=q+2. In this case, q-p=q-2, q, q+2 are three consecutive odd numbers, so one of them must be divisible by 3 (considering the remainders of q divided by 3). The latter is only possible when q-2=3, that is, q=5, r=7.Grading criteria. Only answers: 2 points. Proof that there are no other answers: 5 points.

🧠:[<think>Okay, let's tackle this probability problem. So, we have three people: A, B, and C. Each one of them tries to decrypt a password, and each has a 1/4 chance of succeeding. We need to find the probability that exactly two of them successfully decrypt the password. Hmm, okay.First, let me make sure I understand the problem correctly. Each person acts independently, right? So the success of one doesn't affect the others. That means the probability of any combination of successes and failures can be calculated by multiplying their individual probabilities. Good.We need exactly two successes. So, the possible scenarios are:1. A and B succeed, C fails.2. A and C succeed, B fails.3. B and C succeed, A fails.Each of these is a distinct event, and since they can't happen at the same time (they're mutually exclusive), we can add their probabilities together to get the total probability of exactly two successes.Let me calculate the probability for one scenario first. Let's take the first one: A and B succeed, C fails.The probability that A succeeds is 1/4. Similarly, the probability that B succeeds is 1/4. The probability that C fails is 1 - 1/4 = 3/4. Since they're independent, we multiply these probabilities:(1/4) * (1/4) * (3/4) = (1*1*3)/(4*4*4) = 3/64.Okay, that's 3/64 for one such combination. Now, since there are three different combinations (as listed above), each with the same probability, we can multiply this result by 3.3 * (3/64) = 9/64.So, the probability of exactly two people decrypting the password is 9/64. Wait, let me verify that again. Each combination has two successes and one failure. Each success is 1/4, each failure is 3/4. So each combination's probability is (1/4)^2 * (3/4)^1. Then, multiply by the number of combinations, which is C(3,2) = 3. Therefore, 3 * (1/4)^2 * (3/4)^1 = 3 * 1/16 * 3/4 = 3 * 3/64 = 9/64. Yeah, that seems right.Alternatively, I can think of this as a binomial probability problem, but since there are three trials (each person is a trial), but wait, actually, each person is a Bernoulli trial with success probability 1/4. So the number of successes follows a binomial distribution with parameters n=3 and p=1/4. The probability of exactly k successes is given by C(n,k) * p^k * (1-p)^(n-k). Here, k=2. So substituting the values:C(3,2) * (1/4)^2 * (3/4)^1 = 3 * (1/16) * (3/4) = 3 * 3/64 = 9/64.Yep, same result. So that's consistent. So whether I think of each case individually and add them up, or use the binomial formula, I get 9/64. That seems to check out.Let me just think again if there's any other way this could be approached. Maybe using permutations or something else, but I think the binomial approach is the standard here. Since each trial is independent and identical, binomial is the way to go. So unless there's some trick or hidden detail I'm missing, 9/64 should be the answer.Wait, the problem says "the probability that exactly two of them successfully decrypt the password". So, just to ensure, if two decrypt it, that's what we want. The rest of the wording seems straightforward. Each has 1/4 chance, independent attempts. So yes, the calculation holds. 9/64 is the probability. I think that's solid.Final AnswerThe probability that exactly two of them successfully decrypt the password is boxed{dfrac{9}{64}}.</think>To find the probability that exactly two of A, B, and C successfully decrypt the password, we consider each possible combination of two successes and one failure. Each person has a probability of (frac{1}{4}) of succeeding and (frac{3}{4}) of failing. The possible scenarios for exactly two successes are:1. A and B succeed, C fails.2. A and C succeed, B fails.3. B and C succeed, A fails.For each scenario, the probability is calculated as:[left(frac{1}{4}right) times left(frac{1}{4}right) times left(frac{3}{4}right) = frac{1}{4} times frac{1}{4} times frac{3}{4} = frac{3}{64}]Since there are three such scenarios, we multiply this probability by 3:[3 times frac{3}{64} = frac{9}{64}]This result can also be derived using the binomial probability formula for (k = 2) successes in (n = 3) trials with success probability (p = frac{1}{4}):[binom{3}{2} times left(frac{1}{4}right)^2 times left(frac{3}{4}right)^1 = 3 times frac{1}{16} times frac{3}{4} = 3 times frac{3}{64} = frac{9}{64}]Thus, the probability that exactly two of them successfully decrypt the password is (boxed{dfrac{9}{64}}).,<think>Okay, so I need to find the probability that exactly two out of A, B, and C successfully decrypt a password. Each of them has a 1/4 chance of success, right? And they're doing this independently. Hmm, let's think about how to approach this.First, when dealing with probabilities of multiple independent events, especially where we want a certain number of successes and failures, combinations might be involved here. Like, the number of ways to choose which two people succeed and the third one fails. Yeah, that makes sense. Because the problem is asking for exactly two successes, not at least two. So it's important to consider both the successes and the failures.So, for each possible pair of people (A and B, A and C, B and C), we can calculate the probability that those two succeed and the third fails. Since each attempt is independent, the probabilities should multiply. Let me verify that.Each success has a probability of 1/4, and each failure has a probability of 1 - 1/4 = 3/4. So for one specific pair, say A and B succeed and C fails, the probability would be (1/4) * (1/4) * (3/4). That's 1/4 * 1/4 * 3/4 = 3/64. But there are multiple such pairs. How many pairs are there? Since there are three people, the number of ways to choose two people who succeed is combination of 3 taken 2 at a time. The combination formula is C(n, k) = n! / (k! (n - k)!). So here, C(3, 2) = 3. So there are three such combinations.Therefore, the total probability is 3 * (1/4)^2 * (3/4)^1. Let me compute that. 3 * (1/16) * (3/4) = 3 * 3 / 64 = 9/64. So the probability should be 9/64.Wait, let me check if I did that right. Each of the three scenarios (A and B succeed, C fails; A and C succeed, B fails; B and C succeed, A fails) has a probability of 3/64, and adding them up gives 9/64. Yeah, that seems correct.Alternatively, I could model this using the binomial coefficient. The probability of exactly k successes in n trials is given by C(n, k) * p^k * (1 - p)^(n - k). In this case, though, each trial is a person attempting decryption. Wait, but here each person is like a Bernoulli trial with success probability 1/4. So, n is 3 (three people), k is 2 (two successes). So, plugging into the formula: C(3, 2) * (1/4)^2 * (3/4)^1 = 3 * 1/16 * 3/4 = 9/64. Yep, same answer.Hmm, so that seems straightforward. Is there any chance I missed something? Let's see. The problem states that each person has a probability of 1/4 of decrypting the password. The key word here is "independently," which they mentioned, so the success of one doesn't affect the others. So, multiplying the probabilities is appropriate here.Another way to check is to list all possible outcomes and sum the probabilities of the desired ones. For three people, each can either succeed (S) or fail (F). We want the cases where exactly two are S and one is F.The possible outcomes are:1. A and B succeed, C fails: S S F2. A and C succeed, B fails: S F S3. B and C succeed, A fails: F S SEach of these has the same probability: (1/4)(1/4)(3/4) = 3/64. So adding them up: 3/64 + 3/64 + 3/64 = 9/64. Yep, that's the same as before. So that seems consistent.Alternatively, if I use the formula for the binomial probability distribution, which is exactly designed for this kind of problem: exactly k successes in n independent trials with probability p. Which is what we did. So that's reassuring.I can't think of any other factors here. The problem seems straightforward. Each person's attempt is independent, so no overlaps or dependencies. The answer is 9/64.Final AnswerThe probability that exactly two of them successfully decrypt the password is boxed{dfrac{9}{64}}.</think>To find the probability that exactly two out of A, B, and C successfully decrypt the password, we consider each person's independent probability of success (1/4) and failure (3/4). We need to calculate the probability for each combination of two successes and one failure. There are ( binom{3}{2} = 3 ) such combinations. For each combination, the probability is ( left( frac{1}{4} right)^2 left( frac{3}{4} right) ).Calculating this for one combination:[left( frac{1}{4} right)^2 left( frac{3}{4} right) = frac{1}{16} times frac{3}{4} = frac{3}{64}]Since there are 3 such combinations, the total probability is:[3 times frac{3}{64} = frac{9}{64}]Thus, the probability that exactly two of them successfully decrypt the password is boxed{dfrac{9}{64}}.]

🔑:Given the problem statement:The probability that exactly two people decrypt the password is C_{3}^{2} left( frac{1}{4} right)^{2} cdot frac{3}{4} = frac{9}{64}.Therefore, the answer is: boxed{frac{9}{64}}.Since the probability of each person decrypting the password is the same, we can use combinations to calculate the probability that meets the condition.This question primarily tests basic knowledge of probability and the concept of categorization, examining the ability to use mathematical knowledge to analyze and solve problems.