🧠:[<think>Alright, so I need to solve this triangle problem. Let me start with part (1): Prove that angle A is twice angle B. The given condition is c - b = 2b cos A. Hmm, okay. Let me recall some trigonometric laws related to triangles. The Law of Cosines comes to mind, as well as the Law of Sines. Maybe those can help here.First, let's note that in triangle ABC, side a is opposite angle A, side b is opposite angle B, and side c is opposite angle C. So, given that c - b = 2b cos A. I need to relate the sides and angles here. Let's see.Law of Cosines says that for any triangle,a² = b² + c² - 2bc cos ASimilarly, Law of Sines says that a/sin A = b/sin B = c/sin C = 2R, where R is the radius of the circumscribed circle. Maybe I can use these two laws to relate the given condition.Given c - b = 2b cos A, perhaps I can express c in terms of b and cos A. Let's rearrange the equation:c = 2b cos A + b = b(2 cos A + 1)So c = b(2 cos A + 1). Let's keep that in mind.Now, from the Law of Sines, we have c / sin C = b / sin B. So,c = b (sin C / sin B)But we also have c = b(2 cos A + 1). Therefore,b(2 cos A + 1) = b (sin C / sin B)Divide both sides by b:2 cos A + 1 = sin C / sin BSo sin C = sin B (2 cos A + 1)But in any triangle, angles sum to π, so C = π - A - B. Therefore,sin C = sin(π - A - B) = sin(A + B)Because sin(π - x) = sin x. So sin(A + B) = sin A cos B + cos A sin B.So, substituting back:sin A cos B + cos A sin B = sin B (2 cos A + 1)Let me write that out:sin A cos B + cos A sin B = 2 cos A sin B + sin BSubtract 2 cos A sin B + sin B from both sides:sin A cos B + cos A sin B - 2 cos A sin B - sin B = 0Simplify terms:sin A cos B - cos A sin B - sin B = 0Factor sin B terms:sin A cos B - sin B (cos A + 1) = 0Hmm, not sure if this is helpful. Let me try another approach. Maybe express everything in terms of angle B, since we need to show A = 2B. Let's suppose A = 2B. Then, we can check if the given condition holds. Wait, but the problem is to prove that A = 2B given the condition, so we need to derive it from the equation.Alternatively, maybe express the sides in terms of angles using the Law of Sines and then substitute into the given equation.From Law of Sines:a / sin A = b / sin B = c / sin CLet me denote this common ratio as 2R. So,a = 2R sin Ab = 2R sin Bc = 2R sin CGiven c - b = 2b cos A, substitute:2R sin C - 2R sin B = 2 * 2R sin B * cos ADivide both sides by 2R:sin C - sin B = 2 sin B cos ASo, sin C = sin B + 2 sin B cos A = sin B (1 + 2 cos A)But again, since C = π - A - B, so sin C = sin(A + B). Let me substitute that:sin(A + B) = sin B (1 + 2 cos A)Again, expand sin(A + B):sin A cos B + cos A sin B = sin B (1 + 2 cos A)Bring all terms to one side:sin A cos B + cos A sin B - sin B - 2 sin B cos A = 0Simplify:sin A cos B - sin B - sin B cos A = 0Factor sin B:Wait, let's see term by term:First term: sin A cos BSecond term: - sin BThird term: - sin B cos ASo:sin A cos B - sin B(1 + cos A) = 0Hmm. Let me factor out sin B? Not directly. Alternatively, let's express sin A as sin(2B) if A = 2B. But we need to prove that A = 2B. Let's suppose A = 2B and see if the equation holds. If that's the case, then substituting A = 2B into the equation should satisfy it.So, assuming A = 2B, then let's check:Left side: sin A cos B - sin B(1 + cos A) = sin(2B) cos B - sin B(1 + cos 2B)Compute sin(2B) cos B = 2 sin B cos B * cos B = 2 sin B cos² BCompute 1 + cos 2B = 1 + 2 cos² B - 1 = 2 cos² BSo, sin B(1 + cos 2B) = sin B * 2 cos² BTherefore, the left side becomes:2 sin B cos² B - 2 sin B cos² B = 0Which satisfies the equation. Therefore, if A = 2B, the equation holds. However, we need to ensure that this is the only solution. Suppose there's another possibility. Let's see.Suppose that A ≠ 2B, can the equation still hold? Let's suppose that A = something else. Let me think.Alternatively, maybe express sin A cos B - sin B(1 + cos A) = 0 as:sin A cos B = sin B(1 + cos A)Divide both sides by sin B (assuming sin B ≠ 0, which it can't be in a triangle):(sin A / sin B) cos B = 1 + cos AFrom Law of Sines, sin A / sin B = a / b. But maybe that's not helpful here. Alternatively, using the identity sin A = 2 sin (A/2) cos (A/2). Hmm, not sure.Wait, another approach. Let's divide both sides by cos B:sin A = [sin B(1 + cos A)] / cos BHmm. Let me write this as:sin A = sin B / cos B (1 + cos A) = tan B (1 + cos A)But I don't know if that helps. Alternatively, perhaps use substitution for cos A in terms of angle B. Wait, if we can relate angles A and B, but we need to find that relation.Alternatively, express A in terms of B. Let me denote angle B as x, then angle A would be 2x (if the result is correct). Then, angle C = π - A - B = π - 3x. Let's see if this satisfies the original equation.So, suppose A = 2x, B = x, then C = π - 3x. Then sides:From Law of Sines:a / sin 2x = b / sin x = c / sin(π - 3x) = c / sin 3xTherefore, c = b * sin 3x / sin xSimplify sin 3x = 3 sin x - 4 sin³ x. So,c = b * (3 sin x - 4 sin³ x) / sin x = b(3 - 4 sin² x)Similarly, 2b cos A + b = 2b cos 2x + b = 2b (1 - 2 sin² x) + b = 2b - 4b sin² x + b = 3b - 4b sin² xWhich matches c = 3b - 4b sin² x. Therefore, c = 2b cos A + b. Hence, this holds when A = 2x = 2B. Therefore, this shows that if A = 2B, the given condition holds. But we need to prove the converse: if the given condition holds, then A = 2B. So, the above shows that the condition is satisfied when A = 2B, but does it show that it's the only possibility?Suppose that the equation holds for some A and B. Let's see if there can be another solution. Let's take the equation:sin A cos B - sin B(1 + cos A) = 0Let me rearrange:sin A cos B = sin B(1 + cos A)Divide both sides by sin B cos B (assuming sin B ≠ 0, cos B ≠ 0, which is valid in a triangle):(sin A / sin B) = (1 + cos A) / cos BFrom Law of Sines, sin A / sin B = a / b. But maybe we can use trigonometric identities here.Let me consider writing sin A / sin B = (1 + cos A) / cos BLet me set k = A / B. Suppose k is some constant. If k = 2, then A = 2B is a solution. But perhaps there's another k.Alternatively, express sin A in terms of sin B and cos A in terms of cos B.Wait, perhaps use the formula for sin A in terms of angle B. Let me consider A = 2B. Then sin A = sin 2B = 2 sin B cos B. Let's check:Left side: sin A cos B = 2 sin B cos B * cos B = 2 sin B cos² BRight side: sin B(1 + cos A) = sin B (1 + cos 2B) = sin B (2 cos² B) = 2 sin B cos² BTherefore, equality holds. Now, suppose that A ≠ 2B. Let's see if equality can still hold.Suppose A = 3B. Let's test this. Then, angle C = π - A - B = π - 4B. Then,Left side: sin A cos B = sin 3B cos B = (3 sin B - 4 sin³ B) cos BRight side: sin B(1 + cos A) = sin B(1 + cos 3B) = sin B [1 + 4 cos³ B - 3 cos B]Hmm, let's compute both sides.Left side: (3 sin B - 4 sin³ B) cos B = 3 sin B cos B - 4 sin³ B cos BRight side: sin B [1 + 4 cos³ B - 3 cos B] = sin B [1 - 3 cos B + 4 cos³ B]= sin B - 3 sin B cos B + 4 sin B cos³ BComparing left and right:Left: 3 sin B cos B - 4 sin³ B cos BRight: sin B - 3 sin B cos B + 4 sin B cos³ BSo, set equal:3 sin B cos B - 4 sin³ B cos B = sin B - 3 sin B cos B + 4 sin B cos³ BBring all terms to left:3 sin B cos B - 4 sin³ B cos B - sin B + 3 sin B cos B - 4 sin B cos³ B = 0Combine like terms:(3 sin B cos B + 3 sin B cos B) + (-4 sin³ B cos B) + (- sin B) + (-4 sin B cos³ B) = 0Which is:6 sin B cos B - 4 sin³ B cos B - sin B - 4 sin B cos³ B = 0Factor sin B:sin B [6 cos B - 4 sin² B cos B - 1 - 4 cos³ B] = 0Since sin B ≠ 0, we can divide by sin B:6 cos B - 4 sin² B cos B - 1 - 4 cos³ B = 0Replace sin² B with 1 - cos² B:6 cos B - 4 (1 - cos² B) cos B - 1 - 4 cos³ B = 0Expand:6 cos B - 4 cos B + 4 cos³ B - 1 - 4 cos³ B = 0Simplify:(6 cos B - 4 cos B) + (4 cos³ B - 4 cos³ B) - 1 = 0Which is:2 cos B - 1 = 0Therefore, cos B = 1/2, so B = π/3, but if A = 3B, then A = π, which is impossible in a triangle. So, this would not be a valid solution. Therefore, in this case, the only valid solution would be when A = 2B. Therefore, the equation holds only when A = 2B. Hence, proved.Alternatively, perhaps there is a more straightforward algebraic way. Let's go back to the equation:sin A cos B = sin B(1 + cos A)Divide both sides by sin B:(sin A / sin B) cos B = 1 + cos AFrom the Law of Sines, sin A / sin B = a / b. However, unless we relate this to sides, maybe not helpful here. Alternatively, use trigonometric identities.Let me write sin A as sin(2B) if A = 2B, but again, that's assuming what we need to prove. Maybe use the formula for sin A in terms of angle B. Wait, perhaps express sin A and cos A in terms of angle B.Let me consider the equation:sin A cos B = sin B + sin B cos ARearranged as:sin A cos B - sin B cos A = sin BWhich is:sin(A - B) = sin BBecause sin(A - B) = sin A cos B - cos A sin BTherefore, sin(A - B) = sin BThus, this implies that either:1. A - B = B + 2πk, or2. A - B = π - B + 2πkfor some integer k.But since angles in a triangle are between 0 and π, we can ignore the 2πk terms as k=0.Case 1: A - B = B => A = 2BCase 2: A - B = π - B => A = πBut A = π would mean that the triangle is degenerate, which is impossible. Therefore, the only valid solution is A = 2B. Hence, proved.Ah, that seems much simpler. Using the sine subtraction formula to recognize sin(A - B) = sin B, leading directly to A - B = B, hence A = 2B. That's elegant. I should have thought of that identity earlier.Okay, so part (1) is proved. Now part (2): If cos B = 3/4 and c = 5, find the area of triangle ABC.Given that cos B = 3/4, and c = 5, which is the side opposite angle C. Since we know from part (1) that A = 2B, so angles are A = 2B, B = B, and C = π - A - B = π - 3B.We can use Law of Sines or Law of Cosines to find the sides. Let's see.First, let's note that cos B = 3/4, so sin B = sqrt(1 - (3/4)^2) = sqrt(1 - 9/16) = sqrt(7/16) = sqrt7/4.Since A = 2B, angle A is 2B, so sin A = sin 2B = 2 sin B cos B = 2*(sqrt7/4)*(3/4) = (6 sqrt7)/16 = (3 sqrt7)/8.Angle C is π - 3B. So sin C = sin(3B). Let's compute sin(3B):sin(3B) = 3 sin B - 4 sin^3 B = 3*(sqrt7/4) - 4*(sqrt7/4)^3Compute term by term:First term: 3*(sqrt7/4) = (3 sqrt7)/4Second term: 4*( (sqrt7)^3 / 64 ) = 4*(7 sqrt7 / 64 ) = (28 sqrt7)/64 = (7 sqrt7)/16So sin(3B) = (3 sqrt7)/4 - (7 sqrt7)/16 = (12 sqrt7 - 7 sqrt7)/16 = (5 sqrt7)/16So sin C = (5 sqrt7)/16Now, from Law of Sines:a / sin A = b / sin B = c / sin CGiven c = 5, sin C = 5 sqrt7 /16Thus, the common ratio is 5 / (5 sqrt7 /16 ) = 16 / sqrt7 = (16 sqrt7)/7Therefore:a = (16 sqrt7 /7) * sin A = (16 sqrt7 /7) * (3 sqrt7 /8 ) = (16 sqrt7 * 3 sqrt7 ) / (7*8) = (48*7)/56 = 336 / 56 = 6Similarly,b = (16 sqrt7 /7) * sin B = (16 sqrt7 /7) * (sqrt7 /4 ) = (16*7)/28 = 112 /28 = 4So sides a = 6, b = 4, c =5Now, to compute the area, can use several methods. One is Heron's formula. Compute semi-perimeter:s = (a + b + c)/2 = (6 + 4 +5)/2 =15/2 =7.5Area = sqrt( s(s -a)(s -b)(s -c) ) = sqrt(7.5*(7.5 -6)(7.5 -4)(7.5 -5)) = sqrt(7.5*1.5*3.5*2.5)Hmm, compute this:7.5 *1.5 =11.253.5 *2.5=8.75Then 11.25 *8.75 = let's compute:11 *8 =88, 11*0.75=8.25, 0.25*8=2, 0.25*0.75=0.1875Wait, maybe better to compute 11.25 *8.75:11.25 *8 =9011.25 *0.75 =8.4375So total is90 +8.4375=98.4375Therefore, area = sqrt(98.4375) = sqrt(98.4375). Hmm, but this might not be the simplest way.Alternatively, use area = (1/2)ab sin C. Let's pick sides a and b with angle C between them? Wait, but in triangle ABC, sides a, b, c are opposite angles A, B, C. So, angle C is between sides a and b. Therefore, area = (1/2)ab sin C.Yes, that's straightforward.Compute:Area = (1/2)*a*b*sin C = (1/2)*6*4*(5 sqrt7 /16 )Compute:(1/2)*6*4 =1212*(5 sqrt7 /16 ) = (60 sqrt7)/16 = (15 sqrt7)/4So area is 15 sqrt7 /4Alternatively, using another formula. Let's verify with Heron's formula:sqrt(98.4375). Let's compute 98.4375:Note that 15 sqrt7 /4 ≈15*2.6458/4 ≈39.687/4≈9.9218Compute sqrt(98.4375): sqrt(98.4375) ≈9.9218, so that matches. Therefore, the area is 15√7 /4.Hence, the answer is 15√7 /4, which can be written as boxed{dfrac{15sqrt{7}}{4}}.Final AnswerThe area of ( triangle ABC ) is boxed{dfrac{15sqrt{7}}{4}}.</think>Given a triangle ( triangle ABC ) with sides opposite to angles ( A ), ( B ), and ( C ) being ( a ), ( b ), and ( c ) respectively, we need to prove that ( A = 2B ) given ( c - b = 2b cos A ), and find the area of the triangle when ( cos B = frac{3}{4} ) and ( c = 5 ). Part (1): Prove that ( A = 2B )Starting with the given equation ( c - b = 2b cos A ), we use the Law of Sines and the Law of Cosines. From the Law of Sines, we know:[frac{a}{sin A} = frac{b}{sin B} = frac{c}{sin C}]Given ( c = b(2 cos A + 1) ), substituting into the Law of Sines gives:[sin C = sin B (2 cos A + 1)]Since ( C = pi - A - B ), we use the sine addition formula:[sin(A + B) = sin B (2 cos A + 1)]Expanding and simplifying, we find:[sin A cos B - cos A sin B = sin B]Using the sine subtraction formula, this simplifies to:[sin(A - B) = sin B]This implies ( A - B = B ) (since the other solution would make ( A = pi ), which is impossible), leading to ( A = 2B ). Part (2): Find the area of ( triangle ABC )Given ( cos B = frac{3}{4} ) and ( c = 5 ), we find ( sin B = frac{sqrt{7}}{4} ). Using ( A = 2B ), we calculate:[sin A = sin 2B = 2 sin B cos B = frac{3sqrt{7}}{8}][sin C = sin(3B) = 3 sin B - 4 sin^3 B = frac{5sqrt{7}}{16}]Using the Law of Sines:[frac{a}{sin A} = frac{b}{sin B} = frac{c}{sin C} = frac{16}{sqrt{7}}]We find ( a = 6 ) and ( b = 4 ). The area is calculated using:[text{Area} = frac{1}{2} ab sin C = frac{1}{2} times 6 times 4 times frac{5sqrt{7}}{16} = frac{15sqrt{7}}{4}] Final AnswerThe area of ( triangle ABC ) is boxed{dfrac{15sqrt{7}}{4}}.,<think>Okay, let's tackle this geometry problem step by step. First, part (1) asks us to prove that angle A is twice angle B in triangle ABC, given the condition c - b = 2b cos A. Hmm, interesting. Let me recall some trigonometric laws related to triangles. The Law of Cosines comes to mind, as well as the Law of Sines. Maybe those can help here.So, let's start by writing down what we know. In triangle ABC, sides opposite angles A, B, C are a, b, c respectively. The given equation is c - b = 2b cos A. We need to show that A = 2B.First thought: Maybe express c and b in terms of angles using the Law of Sines. The Law of Sines states that a/sin A = b/sin B = c/sin C = 2R, where R is the circumradius. So, perhaps we can express c as (sin C / sin B) * b, but we might need another relation here.Alternatively, using the Law of Cosines. Let me recall that c^2 = a^2 + b^2 - 2ab cos C. But I don't know if that directly helps here. Wait, the given equation involves c - b and cos A. Maybe we can express cos A in terms of sides using the Law of Cosines. Let's try that.By the Law of Cosines, cos A = (b² + c² - a²)/(2bc). So, substituting this into the given equation:c - b = 2b * [(b² + c² - a²)/(2bc)]Simplify the right side: The 2b in the numerator cancels with the 2bc in the denominator, leaving (b² + c² - a²)/cSo, the equation becomes:c - b = (b² + c² - a²)/cMultiply both sides by c to eliminate the denominator:c(c - b) = b² + c² - a²Expanding the left side: c² - bc = b² + c² - a²Subtract c² from both sides: -bc = b² - a²Rearrange terms: a² - b² = bcSo, now we have a² = b² + bc. Hmm, interesting. Let's note that.So, a² = b² + bc. Maybe we can relate this to angles. Let's recall that by the Law of Sines, a/sin A = b/sin B = c/sin C. Let's denote this common ratio as 2R. So, a = 2R sin A, b = 2R sin B, c = 2R sin C.Substituting into the equation a² = b² + bc:(2R sin A)² = (2R sin B)² + (2R sin B)(2R sin C)Simplify both sides:4R² sin² A = 4R² sin² B + 4R² sin B sin CDivide both sides by 4R²:sin² A = sin² B + sin B sin CSo, sin² A - sin² B - sin B sin C = 0. Hmm, not sure if that helps directly. Let's see if we can relate angles. Since in any triangle, angles add up to π radians (180 degrees), so A + B + C = π. Therefore, C = π - A - B. Maybe substituting that into the equation.So, sin C = sin(π - A - B) = sin(A + B). So, the equation becomes:sin² A - sin² B - sin B sin(A + B) = 0Let me compute sin(A + B) using the sine addition formula: sin(A + B) = sin A cos B + cos A sin B. Therefore:sin² A - sin² B - sin B [sin A cos B + cos A sin B] = 0Expanding the terms:sin² A - sin² B - sin B sin A cos B - sin B cos A sin B = 0Simplify each term:First term: sin² ASecond term: - sin² BThird term: - sin A sin B cos BFourth term: - sin² B cos ASo, combining:sin² A - sin² B - sin A sin B cos B - sin² B cos A = 0Hmm, this seems a bit complicated. Maybe there's a better approach here. Let's think again.We had the equation a² = b² + bc. Let's divide both sides by b²:(a/b)² = 1 + (c/b)But from the Law of Sines, a/b = sin A / sin B. Similarly, c/b = sin C / sin B. So:(sin A / sin B)^2 = 1 + (sin C / sin B)Multiply both sides by sin² B:sin² A = sin² B + sin B sin CWait, we came back to the same equation. Maybe we can express sin C in terms of A and B. Since C = π - A - B, so sin C = sin(A + B). Let's substitute that:sin² A = sin² B + sin B sin(A + B)Again, same as before. Let me use the sine addition formula on sin(A + B):sin(A + B) = sin A cos B + cos A sin B. So:sin² A = sin² B + sin B [sin A cos B + cos A sin B]Which leads to:sin² A = sin² B + sin A sin B cos B + sin B cos A sin BRearranged:sin² A - sin² B - sin A sin B cos B - sin² B cos A = 0Wait, maybe factor terms here. Let's group terms with sin² A and the rest:sin² A - sin A sin B cos B - sin² B (1 + cos A) = 0Not sure if that's helpful. Alternatively, maybe express sin² A - sin² B as sin(A - B) sin(A + B). Remember, sin² x - sin² y = sin(x - y) sin(x + y). Let me use that identity.So, sin² A - sin² B = sin(A - B) sin(A + B). Therefore, the equation becomes:sin(A - B) sin(A + B) - sin B [sin A cos B + cos A sin B] = 0But from earlier, substituting back in:sin(A - B) sin(A + B) - [sin A sin B cos B + sin² B cos A] = 0Hmm, maybe not straightforward. Let's see. Since A + B + C = π, and we have A + B = π - C. But not sure.Alternatively, since we need to prove that A = 2B, perhaps assume that A = 2B and check if the equation holds? But the problem asks us to prove that A = 2B given the condition, so we can't assume it. However, maybe we can use substitution.Suppose A = 2B. Let's check if the equation holds. If that's the case, then C = π - A - B = π - 3B. Let's substitute A = 2B into the equation c - b = 2b cos A.First, express c and b in terms of angles using Law of Sines. c = (sin C / sin B) * b. Since C = π - 3B, sin C = sin(π - 3B) = sin 3B. So, c = (sin 3B / sin B) * b.Therefore, c - b = [ (sin 3B / sin B) - 1 ] * b.Compute sin 3B: sin 3B = 3 sin B - 4 sin³ B. Therefore:[ (3 sin B - 4 sin³ B)/sin B - 1 ] * b = [3 - 4 sin² B - 1] * b = [2 - 4 sin² B] * b.On the other hand, 2b cos A = 2b cos 2B. Since A = 2B, cos A = cos 2B = 1 - 2 sin² B. Therefore, 2b cos A = 2b (1 - 2 sin² B) = 2b - 4b sin² B.Comparing both sides:Left side: [2 - 4 sin² B] * bRight side: 2b - 4b sin² BThey are equal. Therefore, if A = 2B, the equation holds. But does this mean that A must be 2B? Hmm, this shows that A = 2B is a solution, but we need to prove it's the only solution.Alternatively, let's go back to the equation we had: a² = b² + bc. If we can express this in terms of angles and show that A must be 2B.Alternatively, let's use the formula for a² in terms of angles. From the Law of Cosines, a² = b² + c² - 2bc cos A. But we also have a² = b² + bc. Therefore, equate these:b² + c² - 2bc cos A = b² + bcSubtract b² from both sides:c² - 2bc cos A = bcBring bc to the left:c² - 2bc cos A - bc = 0Factor c:c(c - 2b cos A - b) = 0But c ≠ 0, so:c - 2b cos A - b = 0 → c - b = 2b cos AWait, that's the original equation. So, this didn't help. Maybe another approach.Alternatively, let's use the Law of Sines. Let me denote sin A / a = sin B / b = sin C / c = 1/(2R). So, sin A = a/(2R), sin B = b/(2R), sin C = c/(2R).From the equation a² = b² + bc, substitute a = 2R sin A, b = 2R sin B, c = 2R sin C:(2R sin A)^2 = (2R sin B)^2 + (2R sin B)(2R sin C)Divide both sides by (2R)^2:sin² A = sin² B + sin B sin CAs before. Then, since C = π - A - B, so sin C = sin(A + B):sin² A = sin² B + sin B sin(A + B)Let me expand sin(A + B):sin(A + B) = sin A cos B + cos A sin BTherefore:sin² A = sin² B + sin B (sin A cos B + cos A sin B)Expanding the right side:sin² B + sin A sin B cos B + sin² B cos ASo, sin² A = sin² B + sin A sin B cos B + sin² B cos ALet me bring all terms to the left side:sin² A - sin² B - sin A sin B cos B - sin² B cos A = 0Hmm, this seems complicated, but maybe factor terms with sin A and sin B.Alternatively, let's divide both sides by sin² B to simplify terms. Let's set k = A/B, but maybe not helpful. Wait, if we assume A = 2B, does that satisfy the equation? Let's check.Assume A = 2B. Then, sin A = sin 2B = 2 sin B cos B.Left side: sin² A - sin² B - sin A sin B cos B - sin² B cos ASubstitute sin A = 2 sin B cos B, cos A = cos 2B = 2 cos² B - 1.Compute each term:sin² A = (2 sin B cos B)^2 = 4 sin² B cos² Bsin² B = sin² Bsin A sin B cos B = (2 sin B cos B) sin B cos B = 2 sin² B cos² Bsin² B cos A = sin² B (2 cos² B - 1)Putting all into the equation:4 sin² B cos² B - sin² B - 2 sin² B cos² B - sin² B (2 cos² B - 1) = 0Simplify term by term:First term: 4 sin² B cos² BSecond term: - sin² BThird term: - 2 sin² B cos² BFourth term: - sin² B (2 cos² B - 1) = -2 sin² B cos² B + sin² BCombine all terms:4 sin² B cos² B - sin² B - 2 sin² B cos² B -2 sin² B cos² B + sin² BNow, combine like terms:For sin² B cos² B:4 sin² B cos² B - 2 sin² B cos² B -2 sin² B cos² B = 0For sin² B:- sin² B + sin² B = 0So total is 0. Therefore, the equation holds when A = 2B. So that's consistent.But does this mean that A must equal 2B? Or could there be another solution?Suppose there's another solution where A ≠ 2B. Let's assume that A ≠ 2B and see if a contradiction arises.Alternatively, let's try to manipulate the equation sin² A = sin² B + sin B sin C. Let's use C = π - A - B again, so sin C = sin(A + B). Then:sin² A = sin² B + sin B sin(A + B)Expressed as:sin² A - sin² B = sin B sin(A + B)Left side: sin(A - B) sin(A + B) [using identity sin² x - sin² y = sin(x - y) sin(x + y)]Therefore:sin(A - B) sin(A + B) = sin B sin(A + B)Assuming sin(A + B) ≠ 0, which is true because A + B = π - C, and C is an angle in a triangle, so 0 < C < π, so 0 < A + B < π, hence sin(A + B) > 0.Therefore, we can divide both sides by sin(A + B):sin(A - B) = sin BSo, sin(A - B) = sin B. Let's solve this equation.The equation sin θ = sin φ implies θ = φ + 2πn or θ = π - φ + 2πn for some integer n. But since A and B are angles in a triangle, they must be between 0 and π, so n = 0.Therefore, either:1) A - B = B ⇒ A = 2Bor2) A - B = π - B ⇒ A = πBut A = π is impossible in a triangle since the sum of angles is π, so the other angles would have to be zero, which is invalid. Therefore, the only solution is A = 2B.Thus, we have proven that A = 2B.Okay, that was part (1). Now, part (2) asks: If cos B = 3/4 and c = 5, find the area of triangle ABC.Given that A = 2B (from part 1), and cos B = 3/4, c = 5. Let's use this information to find the area.First, since A = 2B, and angles in a triangle sum to π, we have:A + B + C = πSubstitute A = 2B:2B + B + C = π ⇒ 3B + C = π ⇒ C = π - 3BSo, angle C is π - 3B.We know cos B = 3/4, so we can find sin B. Since sin² B + cos² B = 1, sin B = sqrt(1 - (3/4)^2) = sqrt(1 - 9/16) = sqrt(7/16) = √7 / 4.Similarly, since A = 2B, we can find cos A = cos 2B. Using the double-angle formula:cos 2B = 2 cos² B - 1 = 2*(9/16) - 1 = 18/16 - 1 = 2/16 = 1/8.So, cos A = 1/8, and sin A = sin 2B = 2 sin B cos B = 2*(√7/4)*(3/4) = 6√7 / 16 = 3√7 / 8.Now, using the Law of Sines. Let's denote the sides as a, b, c. Given c = 5. From the Law of Sines:a / sin A = b / sin B = c / sin CWe need to find the area. The area of a triangle can be calculated using (1/2)ab sin C or (1/2)bc sin A, etc. But maybe first, let's find the other sides.But we need to find sin C. Since C = π - 3B, sin C = sin(3B). Let's compute sin 3B.Using the identity sin 3B = 3 sin B - 4 sin³ B.We already have sin B = √7 / 4. Compute sin³ B = (√7 / 4)^3 = 7√7 / 64.Therefore, sin 3B = 3*(√7 / 4) - 4*(7√7 / 64) = (3√7)/4 - (28√7)/64 = (3√7)/4 - (7√7)/16 = (12√7 - 7√7)/16 = (5√7)/16.So, sin C = 5√7 / 16.Now, from the Law of Sines:c / sin C = 5 / (5√7 / 16) = 5 * (16 / 5√7) ) = 16 / √7 = (16√7) / 7.Therefore, the common ratio is (16√7)/7. Therefore:a = sin A * (16√7 /7 ) = (3√7 /8 ) * (16√7 /7 ) = (3√7 * 16√7 ) / (8*7)Compute numerator: 3*16 = 48; √7*√7 = 7; so 48*7 = 336Denominator: 8*7 = 56So, a = 336 / 56 = 6.Similarly, b = sin B * (16√7 /7 ) = (√7 /4 ) * (16√7 /7 ) = (√7 *16√7 ) / (4*7 )Numerator: 16*7 = 112 (since √7*√7=7)Denominator: 4*7=28So, b = 112 /28 = 4.Therefore, sides a=6, b=4, c=5.Now, compute the area. Using Heron's formula maybe. Let's check.First, compute semi-perimeter, s = (a + b + c)/2 = (6 + 4 + 5)/2 = 15/2 = 7.5Area = sqrt( s(s - a)(s - b)(s - c) ) = sqrt(7.5*(7.5 - 6)*(7.5 - 4)*(7.5 -5)) = sqrt(7.5*1.5*3.5*2.5)Hmm, that seems a bit messy. Alternatively, since we have sides and angles, maybe use (1/2)ab sin C.Let's choose two sides and the included angle. For example, sides a and b with included angle C. Wait, but we know angle A, B, C. Alternatively, since we have a, b, c, Heron's formula is possible.Alternatively, use (1/2)ab sin C. Let's compute that.We have a =6, b=4, sin C =5√7 /16.So, area = (1/2)*6*4*(5√7 /16 ) = (12)*(5√7 /16 ) = (60√7)/16 = (15√7)/4.Alternatively, check with Heron's formula:s = 7.5s - a = 7.5 - 6 = 1.5s - b = 7.5 -4 =3.5s - c =7.5 -5=2.5Multiply: 7.5*1.5*3.5*2.5Compute 7.5*1.5 = 11.253.5*2.5 =8.75Then, 11.25*8.75. Let's compute:11 *8 =88, 11*0.75=8.25, 0.25*8=2, 0.25*0.75=0.1875. Wait, maybe better to do fractions.7.5 = 15/2, 1.5 =3/2, 3.5=7/2, 2.5=5/2Multiply: (15/2)*(3/2)*(7/2)*(5/2) = (15*3*7*5)/(2^4) = (15*3*35)/16 = (15*105)/16 = 1575/16Then area = sqrt(1575/16) = sqrt(1575)/4Simplify sqrt(1575). 1575=25*63=25*9*7=225*7. So sqrt(1575)=15√7. Therefore, area =15√7/4. Same as before.So the area is 15√7 /4. Let me confirm that with another method, maybe (1/2)bc sin A.We have b=4, c=5, sin A=3√7/8.So area=(1/2)*4*5*(3√7/8)= (10)*(3√7/8)=30√7/8=15√7/4. Yes, same result.Therefore, the area is 15√7/4.Final AnswerThe area of ( triangle ABC ) is boxed{dfrac{15sqrt{7}}{4}}.</think>Given triangle ( triangle ABC ) with sides opposite to angles ( A ), ( B ), and ( C ) being ( a ), ( b ), and ( c ) respectively, we need to prove that ( A = 2B ) given ( c - b = 2b cos A ), and find the area of the triangle when ( cos B = frac{3}{4} ) and ( c = 5 ). Part (1): Prove ( A = 2B )Starting with the given equation ( c - b = 2b cos A ), we use the Law of Cosines to express ( cos A ):[cos A = frac{b^2 + c^2 - a^2}{2bc}]Substituting this into the given equation:[c - b = 2b left( frac{b^2 + c^2 - a^2}{2bc} right)]Simplifying, we get:[c(c - b) = b^2 + c^2 - a^2 implies c^2 - bc = b^2 + c^2 - a^2 implies a^2 = b^2 + bc]Using the Law of Sines, we express ( a ), ( b ), and ( c ) in terms of angles:[frac{a}{sin A} = frac{b}{sin B} = frac{c}{sin C}]Substituting these into ( a^2 = b^2 + bc ):[(2R sin A)^2 = (2R sin B)^2 + (2R sin B)(2R sin C)]Simplifying, we get:[sin^2 A = sin^2 B + sin B sin C]Since ( C = pi - A - B ), we use ( sin C = sin(A + B) ):[sin^2 A = sin^2 B + sin B sin(A + B)]Expanding ( sin(A + B) ):[sin^2 A = sin^2 B + sin B (sin A cos B + cos A sin B)]Simplifying and solving, we find:[sin(A - B) = sin B implies A - B = B implies A = 2B] Part (2): Find the Area of ( triangle ABC )Given ( cos B = frac{3}{4} ) and ( c = 5 ):- ( sin B = sqrt{1 - left(frac{3}{4}right)^2} = frac{sqrt{7}}{4} )- ( A = 2B ) implies ( cos A = cos 2B = 2cos^2 B - 1 = frac{1}{8} )- ( sin A = sin 2B = 2 sin B cos B = frac{3sqrt{7}}{8} )- ( C = pi - 3B ), so ( sin C = sin 3B = 3 sin B - 4 sin^3 B = frac{5sqrt{7}}{16} )Using the Law of Sines:[frac{c}{sin C} = frac{5}{frac{5sqrt{7}}{16}} = frac{16}{sqrt{7}}]Calculating sides ( a ) and ( b ):[a = 6, quad b = 4]Using the area formula ( frac{1}{2}ab sin C ):[text{Area} = frac{1}{2} times 6 times 4 times frac{5sqrt{7}}{16} = frac{15sqrt{7}}{4}] Final AnswerThe area of ( triangle ABC ) is boxed{dfrac{15sqrt{7}}{4}}.]

🔑:Let's proceed with the two parts of the problem separately:( (1) ) We start by using the given equation ( c - b = 2b cos A ) along with the sine rule ( frac{b}{sin B} = frac{c}{sin C} ) to obtain ( sin C - sin B = 2sin B cos A ). Let's call this equation ( (*) ).Since ( C = pi - A - B ), we have[ sin[pi - (A + B)] - sin B = 2sin B cos A, ]which simplifies to[ sin (A + B) - sin B = 2sin B cos A. ]Applying the sum-to-product identities to ( sin (A + B) ), we get[ sin A cos B + cos A sin B - sin B = 2sin B cos A. ]Rearranging, we find[ sin A cos B - cos A sin B = sin B, ]which implies[ sin (A - B) = sin B. ]Given that ( A ) and ( B ) are interior angles of triangle ( triangle ABC ),( B ) is in the interval ( (0, pi) ), and ( A - B ) is also in ( (0, pi) ).Therefore, either ( A - B = B ) or ( A - B + B = pi ) (which we discard as it implies that ( A = pi ), which is not possible inside a triangle).Thus, we conclude that[ A = 2B. ][boxed{A = 2B}]( (2) ) Knowing that ( cos B = frac{3}{4} ) and taking into account that ( B ) lies in the interval ( (0, pi) ), we can find ( sin B ) as follows:[ sin B = sqrt{1 - cos^2 B} = sqrt{1 - left(frac{3}{4}right)^2} = frac{sqrt{7}}{4}. ]Using ( A = 2B ) we can find ( cos A ) and ( sin A ):[ cos A = cos 2B = 2cos^2 B - 1 = 2 × left(frac{3}{4}right)^2 - 1 = frac{1}{8} ][ sin A = sin 2B = 2sin B cos B = 2 × frac{3}{4} × frac{sqrt{7}}{4} = frac{3}{8}sqrt{7} ]From equation ( (*) ) we deduce that ( sin C ) is[ sin C = sin B + 2sin B cos A = frac{sqrt{7}}{4} + 2 × frac{sqrt{7}}{4} × frac{1}{8} = frac{5}{16}sqrt{7}. ]Now, we use the sine rule again in ( triangle ABC ):[ frac{b}{sin B} = frac{c}{sin C} ]Substituting the known values gives us[ frac{b}{frac{sqrt{7}}{4}} = frac{5}{frac{5}{16}sqrt{7}} ]Solving for ( b ) yields[ b = 4. ]Finally, the area ( S ) of ( triangle ABC ) can be computed using[ S = frac{1}{2}bcsin A = frac{1}{2} × 4 × 5 × frac{3}{8}sqrt{7} = frac{15}{4}sqrt{7}. ][boxed{S = frac{15}{4}sqrt{7}}]

🧠:[<think>Okay, so I need to find the sum of the fractional parts of numbers like 1 + 1/2021, 2 + 2/2021, all the way up to 2020 + 2020/2021. Hmm, fractional part is the part after the decimal point, right? For example, {1.25} is 0.25, and {4/3} is 1/3. So, each term in the sum is the fractional part of k + k/2021 where k goes from 1 to 2020.First, let me write each term more clearly. The k-th term is {k + k/2021}. Let's break that down. The integer part of k + k/2021 would just be k, right? Because k is an integer and k/2021 is less than 1 for all k from 1 to 2020. Because 2020/2021 is less than 1. So, the fractional part would be k/2021. Wait, is that right?Wait, if you have k + k/2021, that's equal to k*(1 + 1/2021) = k*2022/2021. So, each term is k*2022/2021. The fractional part of that. Let me see. For example, take k=1: 1*2022/2021 = 2022/2021 ≈1.00049..., so the fractional part is 2022/2021 -1 = 1/2021. Which matches the example given in the problem: {1 + 1/2021} = 1/2021. Similarly, for k=2: 2*2022/2021 = 4044/2021 ≈2.00197..., so fractional part is 4044/2021 -2 = 2/2021. Wait, same pattern. So, in general, for each k, {k +k/2021} = k/2021. Because k*2022/2021 = k +k/2021, so the integer part is k, fractional part is k/2021.Wait, but if that's the case, then the sum is just the sum from k=1 to 2020 of k/2021. Which would be (1 + 2 + ... +2020)/2021. The sum of the first n integers is n(n+1)/2, so here it would be 2020*2021/2 divided by 2021, which simplifies to 2020/2 = 1010. So, the answer is 1010? But that seems too straightforward. Maybe I'm missing something here.Wait, let me check with k=2020. Let's see, 2020*2022/2021. Let's compute that. 2020*2022 = (2021-1)(2021+1) = 2021^2 -1. So, 2021^2 -1 divided by 2021 is 2021 - 1/2021. So, the number is 2021 - 1/2021. The fractional part is {2021 -1/2021} = {2021 -1/2021} = since 2021 is integer, subtract 1/2021 from it. But fractional part is defined as x - floor(x). So, floor(2021 -1/2021) is 2020, because 2021 -1/2021 is just a little less than 2021. So, fractional part is (2021 -1/2021) -2020 = 1 -1/2021 = 2020/2021. Wait, that's different from k/2021. For k=2020, according to my previous logic, the fractional part would be 2020/2021, which actually matches this. So, even for k=2020, {k +k/2021} = k/2021. So, my initial conclusion seems valid.Wait, but then if each term is k/2021, the sum is sum_{k=1}^{2020} k/2021 = (1/2021) * sum_{k=1}^{2020} k = (1/2021)*(2020*2021)/2 = 2020/2 = 1010. So, the answer is 1010. But is that possible?Wait, but let me check with another k. Let's take k=2021. But wait, the problem goes up to k=2020. So, let's take k=2020. Wait, as above, {2020 +2020/2021} = {2020*(1 +1/2021)} = {2020*2022/2021} = { (2021 -1)(2021 +1)/2021 } = { (2021² -1)/2021 } = {2021 - 1/2021} = 1 -1/2021 = 2020/2021. Which is the same as k/2021 where k=2020. So, same result.Alternatively, take k=1011. Then, k +k/2021 =1011 +1011/2021. The integer part is 1011, and the fractional part is 1011/2021. So, yeah, same as before. So, seems like for all k from 1 to 2020, the fractional part {k +k/2021} is k/2021. Therefore, the sum is sum_{k=1}^{2020} k/2021 = (1 +2 +...+2020)/2021 = [2020*2021/2]/2021 = 2020/2 =1010.But the problem is presented as a competition problem, so maybe there is a trick here. Is there a possibility that for some k, k +k/2021 has an integer part greater than k? Because if k/2021 is added to k, since k/2021 is less than 1, the integer part should still be k. So, fractional part is k/2021. So, seems like the answer is 1010.Wait, but wait. Let me check with k=2021. But k goes up to 2020. If k were 2021, then 2021 +2021/2021 =2022, so the fractional part is 0. But k is only up to 2020. So, no, all terms from k=1 to 2020, the fractional part is k/2021. So, the sum is 1010. Therefore, the answer is boxed{1010}.But let me think again. Suppose that the problem had said {k +k/2021}, but when you write k as an integer plus a fractional part, maybe there is some overlapping? Wait, no. For example, take k=2021. But k is only up to 2020. So, for all k in 1 to 2020, k/2021 is less than 1. So, the number k +k/2021 is equal to k + a fraction less than 1, so the integer part is k, and the fractional part is k/2021. Hence, the fractional part is just k/2021. Therefore, the sum is sum_{k=1}^{2020} k/2021 = (1/2021)(2020*2021/2) = 1010.Alternatively, maybe the problem is written as {k +k/2021} where k is written as a mixed number. Wait, the problem says: {1 1/2021}, which is 1 +1/2021, so the fractional part is 1/2021. Similarly, {2 2/2021} is {2 +2/2021} which is 2/2021. So, yes, the fractional part is k/2021. So, the sum is sum_{k=1}^{2020} k/2021.Therefore, 1010. Hmm.Wait, but in the problem statement, they write {1 1/2021}, {2 2/2021}, etc. So, that's 1 +1/2021, 2 +2/2021, etc. So, each term is k + k/2021, and the fractional part is k/2021. So, sum of k/2021 from 1 to2020. So, yeah, same as (1+2+...+2020)/2021 = (2020*2021/2)/2021=1010.But maybe I'm missing something here? Because sometimes fractional part functions can have cases where adding the fractional parts wraps around. Wait, but in each term here, the fractional part is k/2021, which is less than 1, so adding all these fractional parts would just be the sum of k/2021. So, the total sum is 1010.Alternatively, maybe there's a pattern here where each term pairs with another term to sum to 1. Let's see. For example, k and 2021 -k. Let's check.Take k and 2021 -k. Then, the fractional parts would be k/2021 and (2021 -k)/2021. So, adding them together: k/2021 + (2021 -k)/2021 =1. So, each pair sums to 1. How many such pairs are there?Since we are summing from k=1 to 2020, which is 2020 terms. 2020 is even? 2020 divided by 2 is 1010. So, 1010 pairs, each summing to 1. Therefore, total sum is 1010*1=1010. So, same answer. So, that's another way to see it.Yes, so using the pairing method: each pair k and 2021 -k sums to 1. Since 2021 is the denominator, so for k from 1 to 2020, pairing k with 2021 -k gives 1010 pairs each summing to 1. Therefore, total sum is 1010.So, whether you compute it directly or use pairing, you get 1010. Therefore, the answer is 1010.Final Answerboxed{1010}</think>To find the sum of the fractional parts of the numbers ( k + frac{k}{2021} ) for ( k ) ranging from 1 to 2020, we start by noting that the fractional part of ( k + frac{k}{2021} ) is simply ( frac{k}{2021} ). This is because ( k ) is an integer and ( frac{k}{2021} ) is a fraction less than 1, making the integer part of ( k + frac{k}{2021} ) equal to ( k ) and the fractional part equal to ( frac{k}{2021} ).Thus, the sum we need to compute is:[sum_{k=1}^{2020} left{ k + frac{k}{2021} right} = sum_{k=1}^{2020} frac{k}{2021}]This can be rewritten as:[frac{1}{2021} sum_{k=1}^{2020} k]The sum of the first 2020 integers is given by the formula ( frac{n(n+1)}{2} ) with ( n = 2020 ):[sum_{k=1}^{2020} k = frac{2020 times 2021}{2}]Substituting this back into our expression, we get:[frac{1}{2021} times frac{2020 times 2021}{2} = frac{2020}{2} = 1010]Another way to verify this result is by pairing terms ( k ) and ( 2021 - k ). Each pair ( frac{k}{2021} ) and ( frac{2021 - k}{2021} ) sums to 1. There are 2020 terms, forming 1010 pairs, each summing to 1, leading to a total sum of 1010.Thus, the final answer is:[boxed{1010}],<think>Alright, so I have this problem here where I need to find the sum of the fractional parts of numbers like 1 + 1/2021, 2 + 2/2021, all the way up to 2020 + 2020/2021. The fractional part is denoted by {x}, which means for any real number x, {x} = x - floor(x). For example, {1.25} is 0.25 because you subtract the integer part 1, and {4/3} is 1/3 because 4/3 is 1 + 1/3. First, I need to understand what each term in the sum looks like. Let's take the general term: for each k from 1 to 2020, the term is {k + k/2021}. Let me write that out. So, each term is the fractional part of (k + k/2021). Let's break that down. The number k is an integer, and k/2021 is a fractional part. So when we add them together, it's like k + k/2021 = k(1 + 1/2021) = k*(2022/2021). Wait, is that right? Let me check.Wait, k*(1 + 1/2021) = k + k/2021. So each term is k*(2022/2021). But then the fractional part of that would be {k*(2022/2021)}. Hmm, maybe that's a better way to think about it. Alternatively, since k is an integer, the number k + k/2021 is equal to k + (k mod 2021)/2021. Wait, no, not exactly. Let me think again.Wait, if I have an integer k, and I add k/2021 to it, then the integer part of that number is k plus the integer part of k/2021. But k/2021 is a fraction less than 1 if k is less than 2021, right? For k from 1 to 2020, k is less than 2021, so k/2021 is between 0 and 1. Therefore, the integer part of k + k/2021 is just k, and the fractional part is k/2021. But wait, that's only true when k/2021 is less than 1. But k can be up to 2020. So when k is greater than or equal to 2021, then k/2021 is greater than or equal to 1. Wait, but k goes up to 2020. Let me check: when k=2020, then k/2021 = 2020/2021 which is less than 1, so actually, for all k from 1 to 2020, k/2021 is less than 1. Therefore, when we take k + k/2021, the integer part is k, and the fractional part is k/2021. Therefore, {k + k/2021} = k/2021. Therefore, each term in the sum is k/2021. Therefore, the entire sum is the sum from k=1 to 2020 of k/2021.But wait, that can't be right. Let me verify with an example given in the problem statement. For instance, {1.25} = 0.25, which is 1 + 0.25, so fractional part is 0.25, which is 1/4. Similarly, {4/3} is 1/3. So if we have k + k/n, where k < n, then fractional part is k/n. So for k from 1 to n-1, {k + k/n} = k/n. But in the problem, n is 2021, and k goes up to 2020. So if each term is k/2021, then the sum is sum_{k=1}^{2020} k/2021. Then that sum would be (1 + 2 + ... + 2020)/2021. The numerator is the sum of the first 2020 positive integers, which is (2020)(2021)/2. Therefore, the total sum would be (2020*2021/2)/2021 = 2020/2 = 1010. So is the answer 1010? But wait, the problem is written as {1 + 1/2021} + {2 + 2/2021} + ... + {2020 + 2020/2021}, which would be the sum of k/2021 for k from 1 to 2020. So that seems straightforward, but maybe I'm missing something here. Wait, maybe when k is a multiple of 2021, then k/2021 is an integer, so the fractional part is 0, but in our case, k goes up to 2020, so k is never a multiple of 2021. Therefore, all terms would just be k/2021, so sum is (1 + 2 + ... + 2020)/2021. Then, as I calculated, 2020*2021/2 divided by 2021 is 2020/2 = 1010. So the answer is 1010? Hmm. Let me check with a small example. Let's take n=3 instead of 2021, then the sum would be {1 + 1/3} + {2 + 2/3} = 1/3 + 2/3 = 1. Which is (1+2)/3 = 3/3 = 1. Which is (n-1)/2 = (3-1)/2 = 1. So in this case, the formula holds. Wait, wait, in this case, n=3, sum is (1+2)/3=1, which is (3-1)/2=1. So general formula might be (n-1)/2? Wait, but when n=3, 2021 would be n=3 here. Wait, in that case, 2020/2=1010, which is similar to the (n-1)/2 formula. Wait, but here n=2021, so (2021-1)/2=1010, which is the same as 2020/2=1010. So yes, the answer is 1010. But let me confirm again. Let's take another example. Let's take n=4. Then the sum would be {1 + 1/4} + {2 + 2/4} + {3 + 3/4} = 1/4 + 2/4 + 3/4 = (1+2+3)/4 = 6/4 = 1.5. But (n-1)/2 = 3/2 = 1.5. So that works. So for n=4, answer is 1.5. Therefore, the general formula seems to be (n-1)/2. Therefore, for n=2021, the answer is (2021-1)/2=2020/2=1010. So maybe the answer is 1010. But wait, let me check again with n=2. Then sum is {1 + 1/2}=1/2, which is (2-1)/2=0.5. Which is correct. So the formula holds. Therefore, maybe the answer is indeed 1010. However, wait, in the problem statement, the numbers are written as "1 1/2021", "2 2/2021", etc. Are those mixed numbers? For example, "1 1/2021" is 1 + 1/2021, right? So the fractional part is 1/2021. Similarly, "2 2/2021" is 2 + 2/2021, fractional part 2/2021. So yeah, each term is k + k/2021, fractional part is k/2021. Therefore, sum is sum_{k=1}^{2020} k/2021. Which is (sum_{k=1}^{2020} k)/2021. Sum from 1 to m is m(m+1)/2. But here, m=2020, so sum is 2020*2021/2. Therefore, (2020*2021/2)/2021 = 2020/2=1010. So that's straightforward. Therefore, the answer is 1010. But let me check again for another n. Let's take n=5. Then sum from k=1 to 4 of {k + k/5} = sum_{k=1}^4 k/5 = (1+2+3+4)/5=10/5=2. Which is (5-1)/2=2. Correct. So yeah, the formula works. So I think the answer is 1010. But wait, the problem says up to {2020 + 2020/2021}. So k runs from 1 to 2020, inclusive. Therefore, the number of terms is 2020. So the sum is (1 + 2 + ... + 2020)/2021 = [2020*2021/2]/2021 = 2020/2=1010. So the answer is 1010. But I need to make sure there's no trick here. For example, is there any case where {k + k/2021} isn't just k/2021? For example, if k + k/2021 had an integer part greater than k? Wait, but k is an integer, and k/2021 is less than 1 for k from 1 to 2020. So when you add k (integer) and k/2021 (fraction less than 1), the integer part remains k, and the fractional part is k/2021. Therefore, {k + k/2021}=k/2021. Therefore, all terms are just k/2021. Therefore, sum is straightforward. Therefore, answer is 1010. Hmm, seems too straightforward. Maybe the problem is more complicated? Wait, let me check again. Wait, fractional part of x is x minus the greatest integer less than or equal to x. So if x is k + k/2021, then floor(x) is k, since k/2021 is less than 1. Therefore, {x}=x - floor(x)=k + k/2021 - k= k/2021. Therefore, yeah, {k +k/2021}=k/2021. Therefore, the sum is sum_{k=1}^{2020}k/2021. Therefore, sum is (sum_{k=1}^{2020}k)/2021. Sum_{k=1}^n k is n(n+1)/2. Here n=2020, so sum is 2020*2021/2. Divide by 2021, get 2020/2=1010. Therefore, answer is 1010. So unless there's some mistake in my reasoning, that's the answer. Maybe the problem is trying to trick us by writing the numbers as mixed numbers? But no, in the examples given, {1.25}=0.25, which is the same as {1 + 0.25}=0.25, and {4/3}=1/3, which is {1 + 1/3}=1/3. So in both cases, the fractional part is just the fractional part of the mixed number. Therefore, in our problem, each term is a mixed number with integer part k and fractional part k/2021, so {k +k/2021}=k/2021. Therefore, sum is sum_{k=1}^{2020}k/2021=1010. Therefore, the answer is 1010. I think that's correct.Final Answerboxed{1010}</think>To solve the problem, we need to find the sum of the fractional parts of numbers of the form ( k + frac{k}{2021} ) for ( k ) ranging from 1 to 2020. The fractional part of a number ( x ), denoted by ( { x } ), is ( x - lfloor x rfloor ), where ( lfloor x rfloor ) is the greatest integer less than or equal to ( x ).For each ( k ) from 1 to 2020, the term ( k + frac{k}{2021} ) consists of an integer part ( k ) and a fractional part ( frac{k}{2021} ). Since ( frac{k}{2021} ) is always less than 1 for ( k ) in this range, the fractional part of ( k + frac{k}{2021} ) is simply ( frac{k}{2021} ).Thus, the sum we need to compute is:[sum_{k=1}^{2020} left{ k + frac{k}{2021} right} = sum_{k=1}^{2020} frac{k}{2021}]This can be simplified by factoring out the common denominator ( 2021 ):[frac{1}{2021} sum_{k=1}^{2020} k]The sum of the first 2020 positive integers is given by the formula ( sum_{k=1}^{n} k = frac{n(n+1)}{2} ). Substituting ( n = 2020 ):[sum_{k=1}^{2020} k = frac{2020 times 2021}{2}]Substituting this back into our expression, we get:[frac{1}{2021} times frac{2020 times 2021}{2} = frac{2020}{2} = 1010]Therefore, the final answer is:[boxed{1010}]]

🔑:1010

🧠:[<think>Okay, let me try to figure out this problem. So, we have a geometric sequence {a_n} where all terms are positive. The given information is that a₁ * a₅ = 16, and a = 2. Wait, hold on, the problem says "with a = 2". Hmm, maybe that's a typo or mistranslation? Because in the sequence {a_n}, the first term is usually denoted as a₁. So maybe they meant a₁ = 2? Let me check the original problem again.The problem states: "Given that all terms of the geometric sequence {a_n} are positive and it satisfies a₁a₅=16, with a=2, then the common ratio q= (　　)." Hmm, the "a=2" part is a bit unclear. Maybe it's supposed to be a₁=2? That would make sense because if a₁ is 2, then we can use that to find the common ratio. Alternatively, maybe "a" refers to the arithmetic mean or something else? But in the context of a geometric sequence, "a" usually denotes the first term. Let me assume that it's a typo and they meant a₁=2. Otherwise, if "a" is another term, like maybe the fifth term or something, but since the answer options are numbers like 4, 5/2, 2, 1/2, which are common ratios, so probably we need to find q given a₁=2 and a₁*a₅=16.So, let's proceed with that assumption. If a₁ = 2, then a₅ can be expressed in terms of the common ratio q. In a geometric sequence, each term is the previous term multiplied by q. So, aₙ = a₁ * q^(n-1). Therefore, a₅ = a₁ * q^(5-1) = a₁ * q⁴.Given that a₁ * a₅ = 16, substituting the expressions we have:a₁ * (a₁ * q⁴) = 16Since a₁ is 2, substitute that in:2 * (2 * q⁴) = 16Multiply 2 and 2:4 * q⁴ = 16Divide both sides by 4:q⁴ = 4Now, to solve for q, take the fourth root of both sides. Since all terms are positive, q must be positive, so we don't have to consider negative roots.q = 4^(1/4)But 4 is 2², so:q = (2²)^(1/4) = 2^(2/4) = 2^(1/2) = √2Wait, but √2 isn't one of the answer choices. The options are 4, 5/2, 2, 1/2. Hmm, that's a problem. Did I make a mistake here?Wait, maybe my initial assumption was wrong. Let's double-check. The problem says "with a=2". Maybe "a" here refers to the fifth term? Let me see.If "a=2" is referring to a₅=2, then let's try that approach. Then a₁ * a₅ = a₁ * 2 = 16, which would mean a₁ = 8. Then, a₅ = a₁ * q⁴ = 8 * q⁴ = 2. Solving for q:8 * q⁴ = 2q⁴ = 2/8 = 1/4Then q = (1/4)^(1/4). Again, since q is positive, it's the positive root. 1/4 is 2^(-2), so:q = (2^(-2))^(1/4) = 2^(-2/4) = 2^(-1/2) = 1/√2 ≈ 0.707, which is not among the answer choices either.Hmm, neither of these possibilities gives an answer in the options. So maybe I misunderstood the problem statement. Let me read it again:"Given that all terms of the geometric sequence {a_n} are positive and it satisfies a₁a₅=16, with a=2, then the common ratio q= (　　)."Wait, maybe "a=2" is referring to the common ratio? But then the question is asking for q. If a=2 is the common ratio, then q=2, but that would be answer C. But why would they give the common ratio and ask for it? That doesn't make sense.Alternatively, maybe "a=2" is the name of the sequence? But that's not standard terminology.Wait, maybe "a=2" is a mistranslation or a typo. In Chinese problems, sometimes variables can get mixed up in translation. If the original problem was in Chinese, maybe "a" was supposed to be another term. Wait, but the problem is written in English here.Alternatively, maybe "a=2" is referring to the second term, a₂=2? Let's try that. If a₂=2, then since it's a geometric sequence, a₂ = a₁ * q. So a₁ = a₂ / q = 2 / q. Then, given a₁ * a₅ = 16:a₁ * a₅ = (2/q) * (a₁ * q⁴) = (2/q) * ( (2/q) * q⁴ ) = (2/q) * (2 * q³) = 4 * q²Set that equal to 16:4q² = 16q² = 4q = 2 (since q is positive)Then q=2, which is option C. So that works. So maybe "a=2" was supposed to refer to a₂=2? If that's the case, then the answer is C: 2.But how did the problem state it? It says "with a=2". If "a" is a subscript, like a_2=2, then that would make sense. But the way it's written is "with a=2", which is ambiguous. If the original problem had a subscript, maybe it got lost in translation or formatting. For example, if it's written as "a_2=2", then it's clear. But here, it's written as "a=2". So this is a bit confusing.Alternatively, maybe the problem is in French, where "a" is used for "the second term"? No, in French, it's still usually a₁, a₂, etc. Alternatively, maybe "a" is the name of the sequence, but then "a=2" would mean the entire sequence is equal to 2, which doesn't make sense.Alternatively, maybe there was a formatting issue and "a=2" is supposed to be a₁=2. But as we saw earlier, that leads to q=√2, which isn't an answer choice. So that's a problem.Wait, perhaps the problem was "with a₃=2"? Let's test that. If a₃=2, then a₃ = a₁*q²=2. Then, a₁*a₅ = a₁*(a₁*q⁴) = (a₁)^2 * q⁴. But since a₁ = 2 / q², substitute that in:( (2 / q²)^2 ) * q⁴ = (4 / q⁴) * q⁴ = 4 = 16? Wait, 4=16? That's not possible. So that's not right.Alternatively, if a₄=2. Then a₄ = a₁*q³=2. Then a₁ = 2 / q³. Then a₁*a₅ = (2 / q³) * (a₁*q⁴) = (2 / q³) * ( (2 / q³)*q⁴ ) = (2 / q³) * (2*q) = 4*q / q³ = 4 / q² =16So 4 / q² =16q²=4 /16=1/4q=1/2, which is option D. So that gives q=1/2. But the problem stated "with a=2". If "a=2" refers to a₄=2, then the answer would be D: 1/2. But again, the problem statement is unclear.Alternatively, maybe the problem is mistyped, and instead of "a=2", it should be "a₃=2". Wait, let's check.If a₃=2, then a₃=a₁*q²=2. Then a₁=2/q². Then a₁*a₅=(2/q²)*(a₁*q⁴)= (2/q²)*(2/q² * q⁴)= (2/q²)*(2*q²)=4. But 4 ≠16, so that's not matching.Hmm, this is really confusing. Let's recap.Problem states:All terms of the geometric sequence {a_n} are positive.Satisfies a₁a₅=16.With a=2.Find common ratio q.Options: 4, 5/2, 2, 1/2.Possible interpretations:1. "a=2" is a typo and should be a₁=2. Then as we saw, q⁴=4, so q=√2≈1.414, which isn't an option. So invalid.2. "a=2" refers to a₂=2. Then q=2, which is an option. But how is that written? If it's written as "a=2" instead of "a₂=2", that could be a translation error.3. "a=2" refers to a₄=2, leading to q=1/2. But this requires assuming a₄=2.4. Alternatively, maybe the problem has a different structure. Let's see.Wait, maybe the problem is saying that the geometric sequence {a_n} has all terms positive, satisfies a₁*a₅=16, and has a common ratio q=2, then which of the following is the value of a? But the problem is phrased as "with a=2", then find q. So that doesn't align.Alternatively, maybe "a" is a typo for "a₆=2"? Let's test. If a₆=2, then a₆=a₁*q⁵=2. Then a₁*a₅ = a₁*(a₁*q⁴)=a₁²*q⁴. But a₁=2/q⁵. So substituting:( (2 / q⁵)^2 ) * q⁴ = (4 / q¹⁰) * q⁴ = 4 / q⁶ =16So 4 / q⁶=16 → q⁶=4/16=1/4 → q=(1/4)^(1/6)= (2^(-2))^(1/6)=2^(-1/3)=1/2^(1/3)≈0.793, not an option.Alternatively, maybe the problem is written correctly, but we're missing something.Wait, maybe "a=2" is referring to the term a=2, where "a" is a term in the sequence but without specifying the index. But how? If a term in the sequence is 2, but we don't know which one. Then we have a₁*a₅=16 and there exists some term a_n=2. But that's too vague because multiple terms could be 2 depending on q. For example, if a₁=2, then a₅=16/2=8, which would mean q^4=4, so q=√2. If a₂=2, then a₁=2/q and a₅=2*q³, so a₁*a₅=(2/q)*(2*q³)=4*q²=16 ⇒ q²=4 ⇒ q=2. If a₃=2, then a₁=2/q², a₅=2*q², so a₁*a₅=(2/q²)*(2*q²)=4=16, which is impossible. Similarly, a₄=2 would lead to q=1/2. So depending on which term is 2, we get different q. Since the options include 2 and 1/2, perhaps "a=2" refers to a₂=2 (giving q=2) or a₄=2 (giving q=1/2). But we need more information.But the problem states "with a=2", so maybe "a" is a specific term. If "a" is the first term, then a₁=2. If "a" is the second term, a₂=2. But since the problem is translated, maybe the subscript was lost.Alternatively, in Chinese math problems, sometimes they use a, a_2, a_3, etc., but maybe "a=2" is meant to say that one of the terms is 2. But without knowing which term, it's hard to tell. However, since the answer options are given, and both 2 and 1/2 are present, let's see which one fits.If we assume that a₂=2, then q=2 is the answer. If we assume a₄=2, then q=1/2. How can we decide which one? Let's check the original equation a₁*a₅=16.If a₂=2, then a₁=2/q=2/2=1. Then a₅=a₁*q⁴=1*16=16. So a₁*a₅=1*16=16, which matches. So that works. So q=2 is valid.If a₄=2, then a₁=2/q³=2/( (1/2)^3 )=2/(1/8)=16. Then a₅=a₁*q⁴=16*(1/2)^4=16*(1/16)=1. So a₁*a₅=16*1=16, which also works. So both q=2 and q=1/2 satisfy the equation if different terms are set to 2. Therefore, the problem must specify which term is 2. Since the problem says "with a=2", which is ambiguous, but given the answer options, both C and D are possible. However, in the options given, both 2 and 1/2 are present, but the problem might have intended a specific term.Wait, but maybe there's another way. Let's consider that in the problem statement, the Chinese might have used "a" to denote the second term. In some contexts, people use "a" to denote the first term, but if it's the second term, then "a=2" would mean a₂=2. Alternatively, maybe the problem was supposed to say "a_2=2", but the subscript was lost, leading to "a=2". In that case, the answer would be C: 2. Alternatively, if it's a_4=2, then D:1/2. Without more context, it's hard to say. However, given that the answer options include both, and the problem states "a=2", which is more likely to refer to the first term a₁=2. However, as we saw, a₁=2 leads to q=√2, which isn't an option. So that can't be.Alternatively, maybe there was a miscalculation in the problem's creation. If the problem intended a₁=2 and a₃=16, then a₁*a₃=32, which isn't 16 either. Wait, but let's check. If a₁=8 and a₅=2, then q=1/2. As we saw before, if a₅=2, then a₁=8, and q=1/2. Then a₁*a₅=16, which matches, and q=1/2 is an option. So maybe "a=2" refers to a₅=2. But again, the problem states "with a=2", not "a₅=2".Alternatively, maybe "a" is the fifth term. If "a=2" is the fifth term, then a₅=2. Then, as above, a₁=8, q=1/2. So q=1/2. Therefore, answer D.But how do we know? Since both q=2 and q=1/2 satisfy the equation a₁*a₅=16 with different terms set to 2. The problem is ambiguous as stated. However, in the options given, both C and D are possible. But since the problem is likely from a standardized test, there must be only one correct answer. Therefore, perhaps the problem had a typo, and instead of "a=2", it should be "a₂=2" or "a₅=2". Given that the answer options include 2 and 1/2, both are plausible. However, in Chinese textbooks, sometimes problems use "a" to denote the second term, but I'm not sure. Alternatively, maybe the problem is from the Chinese Gaokao or another exam, and in their terminology, "a" is the second term. Alternatively, perhaps the problem originally had "a_2=2" but the subscript was omitted.Given that in the calculation above, if "a=2" is supposed to be the second term (a₂=2), then q=2 is the answer. If it's the fifth term (a₅=2), then q=1/2 is the answer. Since both are possible, but the answer options are C:2 and D:1/2, we need to find which one is correct.Wait, maybe the problem is stated correctly and there's another way to interpret it. Let's try again.If all terms are positive and it's a geometric sequence, with a₁*a₅=16 and "a=2". If "a" is a term in the sequence, then the term "a" is 2. But which term is it? If "a" is the first term, then a₁=2, leading to q=√2 (not an option). If "a" is the second term, then q=2. If "a" is the fifth term, q=1/2. However, without specifying the position, we can't determine. Therefore, the problem is ambiguous. However, considering that the answer options include 2 and 1/2, and given that in geometric sequences, common ratios can be fractions or integers, but both are possible. However, since the product a₁*a₅=16, which is a square number, maybe the common ratio is a whole number. But 2 is a whole number, and 1/2 is a fraction. But both are possible.Wait, but let's think differently. Maybe "a=2" refers to the arithmetic mean? But no, in a geometric sequence, the arithmetic mean isn't typically referred to as "a". Alternatively, maybe "a" is the product of some terms. Hmm.Alternatively, maybe the problem is mistyped and it's supposed to be a_2 * a_5=16, but written as a₁a₅=16. If it's a₂*a₅=16, then let's see:If a=2 is the common ratio, but no, the question is asking for q.Alternatively, maybe the problem is in Chinese where "a" is a variable name, and they define "a" as 2. For example, maybe "set a=2" as part of the problem, not related to the sequence terms. Then, how does that relate? If "a=2" is a separate condition, maybe the sequence has some relation to this variable a. For example, maybe a₁ = a=2, which brings us back to the first scenario where q=√2, but that's not an option.Alternatively, maybe the problem is saying that the sequence {a_n} has a common ratio q, and there's a variable a=2 involved in the equation a₁*a₅=16. For example, maybe the equation is a₁*a₅=16a, and with a=2, then it's 32. But the problem says "it satisfies a₁a₅=16, with a=2". So maybe the equation is a₁*a₅=16a, and with a=2, it becomes 32. But that's pure speculation.Alternatively, maybe "a=2" is the sum of some terms. For example, "with the sum a=2", but that's not mentioned.Given the ambiguity, but considering that both q=2 and q=1/2 are answer choices, and both can satisfy the equation a₁*a₅=16 depending on which term is set to 2, but since the problem says "all terms are positive and it satisfies a₁a₅=16, with a=2", the most plausible interpretation is that "a=2" refers to the first term a₁=2, even though that leads to q=√2, which isn't an option. However, since that's not possible, perhaps the problem intended "a_2=2", leading to q=2, which is an option. Therefore, given the answer choices, and the likely typo, the answer is C:2.Alternatively, if the problem had a₅=2, then q=1/2. But given that in the problem statement "a=2" is mentioned after stating the condition a₁a₅=16, it's possible that "a=2" refers to a variable used in the equation, but that's not clear. However, given the answer choices, and that both 2 and 1/2 are possible, but more likely that if a₂=2, then q=2, which is a simple ratio, maybe the intended answer is C:2.Alternatively, let's notice that in the problem statement, the equation given is a₁a₅=16. In geometric sequences, a₁a₅ = (a₁)^2 q^4. If we let a₃ be the middle term, then a₃ = a₁q², so a₁a₅ = (a₃)^2. Therefore, if a₁a₅=16, then (a₃)^2=16, so a₃=4. Then, if "a=2" refers to a₃=4, but that's not 2. So that's not helpful.Alternatively, if "a=2" is the third term, a₃=2, then a₁a₅ = (a₃/q²)(a₃*q²)=a₃²=4=16? No, that's not possible.Wait, this is getting too convoluted. Given that the problem is likely from a source where "a=2" refers to the second term, leading to q=2, which is an answer option, and considering that possible mistranslation or typo, I think the intended answer is C:2.But I'm still a bit uncertain because without clear indication of which term is 2, both C and D are possible. However, given that option C is a whole number and often common ratios can be integers, and considering possible errors in problem translation, I'll go with C:2.Final Answerboxed{C}</think>Given that all terms of the geometric sequence ({a_n}) are positive and it satisfies (a_1a_5=16), with (a=2), we need to find the common ratio (q).1. Assumption: The problem states "with (a=2)". Assuming this refers to the second term (a_2=2): - In a geometric sequence, the (n)-th term is given by (a_n = a_1 cdot q^{n-1}). - The second term (a_2 = a_1 cdot q = 2). Therefore, (a_1 = frac{2}{q}).2. Using the given product (a_1 cdot a_5 = 16): - The fifth term (a_5 = a_1 cdot q^4). - Substituting (a_1 = frac{2}{q}) into (a_5), we get (a_5 = frac{2}{q} cdot q^4 = 2q^3). - The product (a_1 cdot a_5 = frac{2}{q} cdot 2q^3 = 4q^2). - Given (4q^2 = 16), solving for (q): [ 4q^2 = 16 implies q^2 = 4 implies q = 2 quad (text{since } q text{ is positive}) ]3. Verification: - If (q = 2), then (a_1 = frac{2}{2} = 1). - The fifth term (a_5 = 1 cdot 2^4 = 16). - The product (a_1 cdot a_5 = 1 cdot 16 = 16), which satisfies the given condition.Thus, the common ratio (q) is (boxed{C}).,<think>Okay, let me see. The problem is about a geometric sequence where all terms are positive. It says that a₁ times a₅ equals 16, and a is 2. Wait, hold on, the problem states "with a=2". Wait, but in the sequence notation, it's {a_n}, so maybe a₁ is 2? Let me check the original problem again.The problem says: "Given that all terms of the geometric sequence {a_n} are positive and it satisfies a₁a₅=16, with a=2, then the common ratio q= (　　)." Hmm, the wording is a bit confusing. It says "with a=2". Maybe "a" here refers to the first term a₁? Because in the sequence, the first term is usually denoted as a₁. So if a=2, then a₁=2. That makes sense. Let me confirm that.So, if a₁=2, then we need to find the common ratio q such that a₁ times a₅ is 16. Since it's a geometric sequence, each term is the previous term multiplied by the common ratio q. So, the nth term of a geometric sequence is a_n = a₁ * q^(n-1). Therefore, a₅ would be a₁ * q^(5-1) = a₁ * q⁴.Given that a₁a₅ = 16, substituting the expression for a₅ gives us a₁*(a₁*q⁴) = 16. Since they told us a=2, which I think is a₁=2, then substituting that in gives 2*(2*q⁴) = 16. Let's compute that.Left side: 2*(2*q⁴) = 4*q⁴. So, 4*q⁴ = 16. Dividing both sides by 4: q⁴ = 4. Then, to solve for q, take the fourth root of both sides. Since all terms are positive, q must be positive. So q = 4^(1/4). 4 is 2 squared, so 4^(1/4) is (2²)^(1/4) = 2^(2/4) = 2^(1/2) = √2. Wait, √2 is approximately 1.414, but the answer choices are 4, 5/2, 2, and 1/2. None of these is √2. Did I do something wrong here?Wait, maybe I misunderstood the problem. Let me check again. The problem states "with a=2". If "a" is the first term, then a₁=2. Then a₁a₅=16. So, as above, 2*(2*q⁴)=16. Then 4*q⁴=16, so q⁴=4, q=√2. But that's not one of the options. The options are 4, 5/2, 2, 1/2. Hmm.Wait, perhaps there's a different interpretation. Maybe "a=2" refers to the common ratio? Wait, the common ratio is denoted as q in the problem. The question is asking for q. So maybe "a=2" is referring to another term in the sequence? Wait, the problem says "it satisfies a₁a₅=16, with a=2". Maybe there's a typo here. Maybe "a=2" is supposed to be a specific term, like a_2=2 or a_5=2? Let me read it again carefully.Original problem: "Given that all terms of the geometric sequence {a_n} are positive and it satisfies a₁a₅=16, with a=2, then the common ratio q= (　　)." The wording is a bit unclear. If "with a=2" is given, but "a" hasn't been defined before. Since the sequence is {a_n}, maybe "a" is a typo and should be a specific term, like a_2=2. Alternatively, maybe "a=2" is the first term, so a₁=2, but then when I calculated, I got q=√2, which is not among the options. So perhaps I made a mistake in the calculation.Wait, let me re-examine the calculation step by step. If a₁=2, then a₅=2*q⁴. Then a₁*a₅=2*(2*q⁴)=4*q⁴=16. So q⁴=16/4=4. Then q=4^(1/4)=√2≈1.414. But the options are 4, 5/2, 2, 1/2. None of these. So that suggests that my initial assumption that a=2 refers to a₁=2 is incorrect.Alternatively, maybe "a=2" is referring to another term. Let's suppose that a=2 refers to a_2=2. Then, since a_2 = a₁*q = 2. Then a₁=2/q. Then, given that a₁*a₅=16. Let's compute that. a₁=2/q, a₅ = a₁*q⁴= (2/q)*q⁴= 2*q³. Then a₁*a₅= (2/q)*(2*q³)= (4*q³)/q=4*q². So 4*q²=16, so q²=4, so q=2. Since all terms are positive, q=2. Then the answer would be C:2. That is one of the options. So maybe the problem had a typo, and "a=2" was supposed to be a_2=2. Let me check if that works.If a_2=2, then a₁=2/q. Then a₁*a₅=16. As above, a₅= a₁*q⁴= (2/q)*q⁴=2*q³. Therefore, a₁*a₅=(2/q)*(2*q³)=4*q²=16. Then q²=4, so q=2. So that works, and gives q=2, which is option C. Then that would be the answer. But the problem states "with a=2". If "a" here is supposed to be a term in the sequence, maybe a_2=2. But since the problem didn't specify which term a=2 refers to, perhaps the problem was mistranslated or had a typo.Alternatively, maybe "a=2" refers to the fifth term? Let's see. If a_5=2, then a₁*a₅= a₁*2=16, so a₁=8. Then, since a_5= a₁*q⁴=8*q⁴=2, so q⁴=2/8=1/4, so q= (1/4)^(1/4)= (2^(-2))^(1/4)=2^(-0.5)=1/√2≈0.707, which is also not among the options. So that's not helpful.Alternatively, if "a=2" refers to the third term, a_3=2. Then a₁=2/q². Then a₁*a₅= (2/q²)*(2*q⁴)=4*q²=16, so q²=4, q=2. So again, q=2. So regardless of which term is given as 2 (a_2, a_3, etc.), as long as it's a term in the middle, we can end up with q=2. But the problem didn't specify which term a=2 is. So this is confusing.Wait, let's read the problem again: "Given that all terms of the geometric sequence {a_n} are positive and it satisfies a₁a₅=16, with a=2, then the common ratio q= (　　)."The problem might have a translation error or a typo. If the original problem is in Chinese, maybe the translation is off. Alternatively, maybe "with a=2" is meant to indicate that the sequence has a term equal to 2, but not specifying which one. However, if that's the case, then we have multiple possibilities.Alternatively, maybe "a=2" is a misstatement, and it's supposed to be "a₁=2". If that's the case, then as I originally calculated, q⁴=4, so q=√2, which is not an option. So that can't be.Alternatively, maybe the problem is saying that the geometric sequence has a_1=2, and a_1*a_5=16. Then, solving for q. Let me check that again.If a₁=2, then a₅=2*q⁴. Then a₁*a₅=2*(2*q⁴)=4*q⁴=16. Therefore, q⁴=4, q=4^(1/4)=√2. But this is not one of the answer choices. So this suggests that either the problem is incorrect, or I'm misinterpreting it.Alternatively, maybe "a=2" is referring to the common ratio. Wait, but the question is asking for the common ratio q. So if a=2 is referring to the common ratio, that would be q=2. Then, but then why would they ask for q if it's given? That doesn't make sense.Wait, maybe there's confusion in notation. In Chinese math problems, sometimes "a" is used for the common ratio. Wait, no, the problem clearly states "the common ratio q". So q is the common ratio. So "a" must refer to a term in the sequence.Alternatively, maybe "a=2" is the sum of some terms? But the problem doesn't specify. Alternatively, maybe "a=2" is a misprint and should be "a_5=2". Let's test that. If a_5=2, then a₁*a₅= a₁*2=16, so a₁=8. Then a_5= a₁*q⁴=8*q⁴=2. So q⁴=2/8=1/4, so q=(1/4)^(1/4)= (2^(-2))^(1/4)=2^(-0.5)=1/√2≈0.707, which again is not an option.Alternatively, if "a=2" refers to a_3=2, then a₁=2/q². Then a₁*a₅=(2/q²)*(2*q⁴)=4*q²=16. Thus, q²=4, so q=2. So this gives q=2. So if "a=2" is supposed to be a_3=2, then the answer is C:2. But the problem didn't specify which term is 2.Alternatively, perhaps "with a=2" is supposed to mean that the sequence has 2 as one of its terms. But since all terms are positive and it's a geometric sequence, unless it's a constant sequence, which would have q=1, but that's not an option. If a=2 is a term in the sequence, but not specified which, then there are multiple possibilities. But given the answer options, maybe we can infer.Alternatively, maybe the problem is mistyped. If the original problem was "a₁=2 and a₅=16", then the common ratio would be q=2, since a₅ = a₁*q⁴ => 16=2*q⁴ => q⁴=8 => q=√8^(1/4) => Hmm, no. Wait, 16=2*q⁴ => q⁴=8 => q=8^(1/4)=2^(3/4), which is approximately 1.68, not an option. Hmm.Alternatively, if the problem was supposed to say that a₁=2 and a₅=16, then solving for q: a₅=2*q⁴=16 => q⁴=8 => q=8^(1/4)=2^(3/4). Not helpful.Alternatively, if the problem was supposed to say that a₁=2 and a₁*a₅=16, then as before, 2*a₅=16 => a₅=8. Then a₅=2*q⁴=8 => q⁴=4 => q=√2. Still not an option.Alternatively, perhaps "a=2" is a misstatement for "a_2=2", which would lead to a₁=2/q, and then a₁*a₅=16. Then, as I did before, leading to q=2. Since this is one of the answer choices, and given that the answer is C:2, which is an option, perhaps the intended interpretation is that a_2=2.But the problem says "with a=2", not "a_2=2". So this is a problem of unclear notation. However, given that the answer options include 2, and through some interpretation, we can get to q=2, which is option C, then maybe that's the intended answer. Alternatively, perhaps the problem statement in Chinese was clearer, but in translation, it's ambiguous.Alternatively, maybe the problem is from a source where "a" is used to denote the second term. But that's non-standard. Normally, a₁ is the first term, a₂ is the second term, etc.Wait, another approach. Let's check all the options. Since it's multiple-choice, maybe we can test each possible q and see which one satisfies the conditions.Given that all terms are positive, and the common ratio q is one of the options: 4, 5/2, 2, or 1/2.The problem states that a₁a₅=16, and "with a=2". If "a=2" is supposed to mean a specific term is 2, maybe a₁=2. Let's test each q:Option C: q=2. If q=2 and a₁=2, then a₅=2*2^(4)=2*16=32. Then a₁a₅=2*32=64≠16. So that's not correct.Wait, that's a problem. If a₁=2 and q=2, then a₅=32, product is 64, not 16. So if q=2 is not the answer under the assumption that a₁=2.Alternatively, if "a=2" refers to a different term. Suppose a₂=2. Then a₁=2/q. Then a₅= a₁*q⁴= (2/q)*q⁴=2*q³. Then a₁a₅= (2/q)*(2*q³)=4*q²=16. Therefore, q²=4, so q=2. So if a₂=2, then q=2. So then the product is 16. But the problem says "with a=2". If "a=2" is a₂=2, then this works, but the problem didn't specify. However, the answer is C:2, which works under this assumption.Alternatively, if "a=2" is a₃=2, then a₁=2/q². Then a₅= a₁*q⁴=2/q² * q⁴=2*q². Then a₁a₅= (2/q²)*(2*q²)=4=16? No, that's 4, not 16. So that doesn't work.Wait, so only if a₂=2, then the product is 16. Therefore, if the problem had specified that a₂=2, then q=2. However, since the problem says "with a=2", and given the answer options, maybe this is the intended path. Therefore, despite the ambiguity, the answer is C:2.Alternatively, maybe the problem was meant to say that the second term is 2, i.e., a_2=2, and a₁a₅=16. Then as above, q=2. Since option C is 2, which is one of the answer choices, and the other options don't fit, then this must be the answer. Therefore, the answer is C:2.Final Answerboxed{C}</think>Given that all terms of the geometric sequence ({a_n}) are positive and it satisfies (a_1a_5=16), with (a=2), we need to find the common ratio (q).Assuming (a=2) refers to the second term (a_2=2), we can determine the common ratio (q). 1. In a geometric sequence, the (n)-th term is given by (a_n = a_1 cdot q^{n-1}).2. The second term (a_2) is (a_1 cdot q = 2). Therefore, (a_1 = frac{2}{q}).3. The fifth term (a_5) is (a_1 cdot q^4). Substituting (a_1 = frac{2}{q}), we get (a_5 = frac{2}{q} cdot q^4 = 2q^3).4. Given (a_1 cdot a_5 = 16), substituting the expressions for (a_1) and (a_5): [ left(frac{2}{q}right) cdot (2q^3) = 16 ]5. Simplifying this equation: [ frac{4q^3}{q} = 16 implies 4q^2 = 16 implies q^2 = 4 implies q = 2 ]Thus, the common ratio (q) is (boxed{C}).]

🔑:AnalysisThis question examines the application of the general formula and properties of a geometric sequence, and it is a basic problem.SolutionGiven (a_1a_5=16) and that all terms are positive, we can deduce (a_3=4),Also, (a_3=a_2q),which means (4=2q),Therefore, (q=2).Hence, the correct choice is boxed{C}.

🧠:[<think>Okay, so I need to figure out whether the function f(x) = lg|sin x| is odd or even, and what its smallest positive period is. The options are A, B, C, D with combinations of odd/even and periods π or 2π. Let me start by recalling what odd and even functions are.An even function satisfies f(-x) = f(x) for all x in its domain. An odd function satisfies f(-x) = -f(x) for all x in its domain. So first, I need to check if f(-x) equals f(x) or -f(x).Given f(x) = lg|sin x|. Let's compute f(-x):f(-x) = lg|sin(-x)| = lg| -sin x | because sin(-x) = -sin x. The absolute value of -sin x is the same as |sin x|. So f(-x) = lg|sin x| = f(x). Therefore, f(-x) = f(x), so the function is even. That rules out options A and B, which say it's odd.So now the answer is either C or D. The next part is determining the period. The function is f(x) = lg|sin x|. Let's recall the period of |sin x| first. The sine function has a period of 2π, but when we take the absolute value, |sin x|, the period becomes π because the negative parts are flipped to positive, so the graph repeats every π instead of 2π.Therefore, |sin x| has a period of π. Then, taking the logarithm of it, does that affect the period? The logarithm is applied to the absolute value of sine, but since the logarithm is a function applied after the absolute sine, the period should remain the same as |sin x|, right? Because applying a function to a periodic function doesn't change its period unless the function itself has some periodicity that interacts with it. But logarithm is not a periodic function, so the period of lg|sin x| should be the same as |sin x|, which is π.Wait, but wait. Let me verify this. Suppose we have a function g(x) = |sin x|, which has period π. Then f(x) = lg(g(x)). So the logarithm is just scaling the values, but the periodicity is determined by how often g(x) repeats. Since g(x) repeats every π, f(x) should also repeat every π. Therefore, the period of f(x) is π. So that would make the answer C: an even function with the smallest positive period of π.But let me check if there could be a smaller period. Suppose there is a period smaller than π. Let's assume T is the period of f(x). Then, for all x, lg|sin(x + T)| = lg|sin x|. Which implies |sin(x + T)| = |sin x| for all x. Since logarithm is injective over positive reals (which is the domain here because |sin x| is non-negative and we can take logarithm except when |sin x|=0, but logarithm is defined only where |sin x| > 0, i.e., when sin x ≠ 0). So where the function is defined, the equality |sin(x + T)| = |sin x| must hold. The question is, what's the smallest positive T for which this is true.We know that |sin x| has period π. Let's check if π is indeed the fundamental period. Suppose there exists a T smaller than π such that |sin(x + T)| = |sin x| for all x. Let's test T = π/2. For x = 0, |sin(0 + π/2)| = |sin(π/2)| = 1, and |sin 0| = 0. So 1 ≠ 0, so T=π/2 is not a period. How about T=π/3? Take x=0: |sin(π/3)| = sqrt(3)/2, but |sin 0|=0. Not equal. Similarly, any T < π would have points where |sin(x + T)| ≠ |sin x|. Therefore, π is indeed the fundamental period of |sin x|, and thus of lg|sin x|.Therefore, the function f(x) is even and has period π. So the correct answer is C.But hold on. Let me check another point. Suppose we take x = π/4. Then sin(π/4) = sqrt(2)/2. Then f(π/4) = lg(sqrt(2)/2). Then if we add the period T, say π, x + T = 5π/4. Then sin(5π/4) = -sqrt(2)/2, absolute value is sqrt(2)/2, so f(5π/4) = lg(sqrt(2)/2) which is equal to f(π/4). So that works. If we tried T=π/2, then x + T = 3π/4, sin(3π/4)=sqrt(2)/2, absolute same, so f(3π/4) = same value. Wait, but if we check another x. Take x=π/6. Then f(π/6)=lg(1/2). If we add T=π/2, x + T= 2π/3, sin(2π/3)=sqrt(3)/2, which is different from sin(π/6)=1/2. So |sin(2π/3)|=sqrt(3)/2 ≠ 1/2, so f(x + T)=lg(sqrt(3)/2) ≠ lg(1/2). Therefore, T=π/2 is not a period. Thus, the period can't be smaller than π. So yes, π is indeed the fundamental period.Therefore, the answer should be C. But let me just make sure I didn't make any mistake in the parity. The function is even because f(-x) = lg|sin(-x)| = lg| - sinx | = lg|sinx| = f(x). So yes, even function. So C is the correct option. D says period 2π, but since the fundamental period is π, so C is correct.Final Answerboxed{C}</think>To determine whether the function ( f(x) = lg|sin x| ) is odd or even and its smallest positive period, we start by checking its parity. 1. Parity Check: - Compute ( f(-x) ): [ f(-x) = lg|sin(-x)| = lg|-sin x| = lg|sin x| = f(x) ] - Since ( f(-x) = f(x) ), the function is even. This eliminates options A and B.2. Periodicity Check: - The function ( |sin x| ) has a period of ( pi ) because the absolute value of the sine function repeats every ( pi ). - Applying the logarithm function to ( |sin x| ) does not change its periodicity because the logarithm is a non-periodic function. Therefore, the period of ( lg|sin x| ) remains ( pi ).3. Verification of Fundamental Period: - To confirm ( pi ) is the smallest period, we check if any smaller ( T ) (e.g., ( pi/2 )) can satisfy ( |sin(x + T)| = |sin x| ) for all ( x ). Testing ( T = pi/2 ) and other values smaller than ( pi ) shows they do not satisfy the condition, confirming ( pi ) is the fundamental period.Thus, the function ( f(x) = lg|sin x| ) is an even function with the smallest positive period of ( pi ).[boxed{C}],<think>Okay, let's try to figure out this problem. The question is about determining whether the function f(x) = lg|sinx| is odd or even, and what its smallest positive period is. The options are A through D, with combinations of odd/even and periods π or 2π. Alright, let's break this down step by step.First, I need to recall the definitions of odd and even functions. An even function satisfies f(-x) = f(x) for all x in its domain, and an odd function satisfies f(-x) = -f(x) for all x in its domain. So, I need to check whether f(-x) equals f(x) or -f(x).The function here is f(x) = lg|sinx|. Let's compute f(-x). Replacing x with -x, we get:f(-x) = lg|sin(-x)|.Now, sin(-x) is equal to -sinx, right? Because sine is an odd function. So sin(-x) = -sinx. Then, taking the absolute value of that, |sin(-x)| = |-sinx| = |sinx|. So the absolute value removes the negative sign. Therefore, |sin(-x)| = |sinx|.So then, f(-x) = lg|sinx| = f(x). Therefore, f(-x) = f(x), which by definition means that the function is even. So that rules out options A and B, which say it's an odd function. So now we're left with options C and D, which both say it's even, differing in the period.Now, we need to find the smallest positive period of the function. The function is f(x) = lg|sinx|. Let's think about the periodicity of sinx. The regular sine function has a period of 2π. However, here we have the absolute value of sinx, |sinx|. The absolute value of sinx would take the negative parts of the sine wave and flip them to positive, effectively creating a waveform that repeats every π instead of 2π. Because if you take sinx from 0 to π, it's non-negative, and from π to 2π, it's non-positive, but taking absolute value would make it the same as 0 to π again. So |sinx| has a period of π.But wait, our function is the logarithm of that, lg|sinx|. The logarithm function is only defined when its argument is positive. However, since we have the absolute value inside the logarithm, |sinx| is always non-negative, so the logarithm is defined whenever |sinx| > 0. That is, except when sinx = 0, which occurs at integer multiples of π. Therefore, the domain of f(x) is all real numbers except x = nπ, where n is an integer. But how does the logarithm affect the periodicity? The logarithm is a monotonically increasing function, but applying it to a periodic function doesn't necessarily change the period. The period of the composition would still be the same as the period of the inner function, provided that the outer function doesn't introduce any new periodicity. Since the logarithm is not periodic, the period of lg|sinx| should be the same as the period of |sinx|, which is π.However, let's verify this. Let's check if f(x + π) = f(x). Compute f(x + π) = lg|sin(x + π)|. Now, sin(x + π) = -sinx, so |sin(x + π)| = |-sinx| = |sinx|. Therefore, lg|sin(x + π)| = lg|sinx| = f(x). Therefore, the function f(x) is periodic with period π. Is π the smallest positive period?Suppose there exists a smaller period, say T < π, such that f(x + T) = f(x) for all x in the domain. Let's test this. Suppose T is π/2. Let's check if f(x + π/2) = f(x). Compute f(x + π/2) = lg|sin(x + π/2)|. sin(x + π/2) = cosx, so this becomes lg|cosx|. Compare this to f(x) = lg|sinx|. So unless lg|cosx| = lg|sinx| for all x, which is not true. For example, take x = 0. Then f(0 + π/2) = lg|cos0| = lg1 = 0, but f(0) = lg|sin0|, which is undefined (since sin0 = 0), but near x=0, as x approaches 0 from the right, sinx approaches 0, so lg|sinx| approaches negative infinity. So near x=0, the function tends to -infty, but at x=π/2, the function is 0. So clearly, shifting by π/2 does not preserve the function's values. Hence, T=π/2 is not a period.How about T=π? As we saw before, f(x + π) = f(x). So π is a period. Now, is there a smaller period than π? Suppose we take T=π/3. Let's test f(x + π/3) = lg|sin(x + π/3)|. Let's pick a specific x to check. Let x=0. Then f(0 + π/3) = lg|sin(π/3)| = lg(√3/2) ≈ lg(0.866). f(0) is undefined, but approaching from the right, it's approaching -infty. So clearly, f(x + π/3) is not equal to f(x) near x=0. So T=π/3 is not a period.Alternatively, take x=π/6. Then f(π/6 + π) = f(7π/6) = lg|sin(7π/6)| = lg| -1/2 | = lg(1/2) = -lg2. On the other hand, f(π/6) = lg|sin(π/6)| = lg(1/2) = -lg2. So that's equal, but that's expected since π is a period. If we take T=π/2, as before, x=π/4. Then f(π/4 + π/2) = f(3π/4) = lg|sin(3π/4)| = lg(√2/2) ≈ -0.1505. f(π/4) = lg|sin(π/4)| = same value, -0.1505. Hmm, so that's equal. Wait, but earlier when I took x=0, f(π/2) was 0, but f(0) is undefined. But maybe there's some overlapping?Wait, maybe I need a different approach. Suppose T is a period of f(x). Then for all x where x is in the domain, x + T must also be in the domain, and f(x + T) = f(x). The original function f(x) = lg|sinx| has domain all real numbers except x = nπ, n integer. So the points of discontinuity are at x = nπ.If T is a period, then the set of discontinuities must also be periodic with period T. That is, if x = nπ is a discontinuity, then x + T = nπ + T must also be a discontinuity. Therefore, T must satisfy that T is a multiple of π, since the discontinuities are spaced π apart. Therefore, the period must be a multiple of π. The fundamental period would then be π, since π is the smallest such positive number for which shifting by T=π maps discontinuities to discontinuities (nπ + π = (n+1)π, which is another discontinuity). If we tried a smaller period, say T=π/k for some integer k >1, then the discontinuities would need to be at x = nπ + mT for integers m. But unless T divides π, this would not hold. For example, if T=π/2, then starting from x=0, adding T would give π/2, which is not a discontinuity. But π/2 is not a point where the function is undefined, since sin(π/2)=1, so f(π/2)=lg1=0. So in that case, the function is defined at π/2, but undefined at π. Therefore, shifting by T=π/2 does not preserve the domain, hence T=π/2 cannot be a period. Therefore, the minimal period is π.Thus, the function is even with period π, which is option C.But wait, let's check once more to make sure. Let's verify with specific examples. Take x=π/4. Then f(π/4)=lg|sin(π/4)|=lg(√2/2)=lg(1/√2)= - (1/2) lg2 ≈ -0.1505. Now, check f(-π/4)=lg|sin(-π/4)|=lg| -√2/2 | = same as f(π/4), so that's equal. So even function confirmed.For the period, check f(x + π)=lg|sin(x + π)|=lg| -sinx | = lg|sinx|=f(x). So period π. Is there a smaller period? Suppose T=π/2. Let's pick x=π/4. Then f(π/4 + π/2)=f(3π/4)=lg|sin(3π/4)|=lg(√2/2)= same as before. But does this hold for all x? Take x=π/6. f(π/6)=lg(1/2). Then f(π/6 + π/2)=f(2π/3)=lg|sin(2π/3)|=lg(√3/2). But lg(1/2) is not equal to lg(√3/2), since 1/2 ≈0.5 and √3/2≈0.866. Therefore, their logarithms are different. Therefore, f(x + π/2) ≠ f(x) in general. So π/2 is not a period. Therefore, π is indeed the minimal period.Therefore, the correct answer should be C: An even function with the smallest positive period of π.Final Answerboxed{C}</think>To determine whether the function ( f(x) = lg|sin x| ) is odd or even and its smallest positive period, we start by checking the parity of the function.1. Check if the function is odd or even: - Compute ( f(-x) ): [ f(-x) = lg|sin(-x)| = lg|-sin x| = lg|sin x| = f(x) ] - Since ( f(-x) = f(x) ), the function is even.2. Determine the periodicity: - The function inside the logarithm is ( |sin x| ), which has a period of ( pi ) because ( |sin(x + pi)| = |-sin x| = |sin x| ). - The logarithm function does not affect the periodicity of ( |sin x| ), so the period of ( f(x) = lg|sin x| ) is also ( pi ).3. Verify the smallest period: - Check if a smaller period ( T < pi ) exists. For example, testing ( T = pi/2 ): - Compute ( f(x + pi/2) = lg|sin(x + pi/2)| = lg|cos x| ), which is not equal to ( lg|sin x| ) for all ( x ). - Therefore, ( pi ) is indeed the smallest period.Thus, the function ( f(x) = lg|sin x| ) is an even function with the smallest positive period of ( pi ).[boxed{C}]]

🔑:Firstly, we recognize that the domain of f(x) precludes values of x that make sin x equal to zero, as the logarithm function is not defined for zero. Therefore, the domain of f(x) is {x | x neq kpi, k in mathbb{Z}}.The function sin x has the property that sin (-x) = -sin x, and since we are taking the absolute value |sin x|, this gives us |sin (-x)| = |sin x|. Therefore, when we apply the logarithmic function, we obtain the following: f(-x) = lg|sin(-x)| = lg|sin x| = f(x), indicating that f(x) is an even function.Next, we need to identify the smallest positive period of f(x). We know that the function y = sin x has a period of 2pi. However, taking the absolute value of sin x modifies the period because |sin x| is symmetrical about the x-axis, effectively halving the original period of the sine function. Thus, y = |sin x| has a period of pi.Since taking the logarithm will not affect the period of the function, we conclude that f(x) = lg|sin x| has the same period as y = |sin x|, which is pi. Therefore, f(x) is an even function with the smallest positive period of pi.[boxed{C}] is the correct choice.

🧠:[<think>Alright, let's see. I need to find the sum of the coefficients of all terms in the expansion of the expression (x³ + 2x + 1)(3x² + 4). Hmm, okay. Let me start by recalling how to expand such expressions. It's been a while, but I think I need to use the distributive property, also known as the FOIL method for binomials, but since both polynomials here have more than two terms, I need to distribute each term in the first polynomial to each term in the second polynomial and then combine like terms. First, let me write down the two polynomials:First polynomial: x³ + 2x + 1Second polynomial: 3x² + 4So, to expand, I have to multiply each term in the first polynomial by each term in the second one and add all the results together. Let me break it down step by step.Multiply x³ by each term in the second polynomial:x³ * 3x² = 3x^(3+2) = 3x⁵x³ * 4 = 4x³Then, multiply 2x by each term in the second polynomial:2x * 3x² = 6x^(1+2) = 6x³2x * 4 = 8xNext, multiply 1 by each term in the second polynomial:1 * 3x² = 3x²1 * 4 = 4Now, let's list all the products we obtained:3x⁵, 4x³, 6x³, 8x, 3x², 4Now, we need to combine like terms. Let's look for terms with the same exponent.First, 3x⁵ is the only term with x⁵.Then, the terms with x³ are 4x³ and 6x³. Adding those together: 4x³ + 6x³ = 10x³.Next, the term with x² is 3x².The term with x is 8x.And the constant term is 4.So putting all the combined terms together, the expanded polynomial is:3x⁵ + 10x³ + 3x² + 8x + 4Now, the question asks for the sum of the coefficients of all terms in the expansion. The coefficients are the numerical parts of each term. Let me list them:- The coefficient of x⁵ is 3- The coefficient of x³ is 10- The coefficient of x² is 3- The coefficient of x is 8- The constant term is 4, which is technically the coefficient of x⁰So, adding all these coefficients together: 3 + 10 + 3 + 8 + 4.Let me compute that step by step:3 + 10 = 1313 + 3 = 1616 + 8 = 2424 + 4 = 28So, the sum of the coefficients is 28.Wait a second, but there's a shortcut to find the sum of coefficients in a polynomial. I remember that the sum of the coefficients can be found by evaluating the polynomial at x = 1. Let me verify that.If I substitute x = 1 into the original expression before expanding, then (1³ + 2*1 + 1)(3*1² + 4) = (1 + 2 + 1)(3 + 4) = (4)(7) = 28. That gives the same result. So that's a good check. Instead of expanding and adding the coefficients, I could have just plugged in x = 1 into the original polynomials and multiplied the results. That would have saved time. But since I had to expand it anyway for the problem, maybe they wanted to see the expanded form first? Well, regardless, both methods give the same answer, which is reassuring.But to make sure I didn't make a mistake in expanding, let me check the expansion again:Original polynomials:(x³ + 2x + 1)(3x² + 4)Multiply x³ by 3x²: 3x^5Multiply x³ by 4: 4x³Multiply 2x by 3x²: 6x³Multiply 2x by 4: 8xMultiply 1 by 3x²: 3x²Multiply 1 by 4: 4Combine like terms:3x^5 + (4x³ + 6x³) + 3x² + 8x + 4Which is 3x^5 + 10x³ + 3x² + 8x + 4. Yep, that's correct.Then coefficients: 3, 10, 3, 8, 4. Sum is 28. That seems right.Alternatively, using the substitution method, plugging in x=1:First polynomial at x=1: 1^3 + 2*1 + 1 = 1 + 2 + 1 = 4Second polynomial at x=1: 3*1^2 + 4 = 3 + 4 = 7Multiply 4*7 = 28. Perfect, same answer.So, whether expanding or substituting, the sum of the coefficients is 28. Therefore, the answer should be 28.Final AnswerThe sum of the coefficients is boxed{28}.</think>To determine the sum of the coefficients of all terms in the expansion of ((x^{3} + 2x + 1)(3x^{2} + 4)), we can use the shortcut method of evaluating the polynomial at (x = 1).First, substitute (x = 1) into each polynomial:- For the first polynomial (x^{3} + 2x + 1): [ 1^{3} + 2 cdot 1 + 1 = 1 + 2 + 1 = 4 ]- For the second polynomial (3x^{2} + 4): [ 3 cdot 1^{2} + 4 = 3 + 4 = 7 ]Next, multiply the results of these substitutions:[4 times 7 = 28]Thus, the sum of the coefficients of all terms in the expansion is (boxed{28}).,<think>Okay, so I need to find the sum of the coefficients of all terms in the expansion of the expression (x³ + 2x + 1)(3x² + 4). Hmm, let me think about how to approach this. First, I remember that when you're asked for the sum of coefficients in a polynomial, there's a shortcut where you can substitute x = 1 into the polynomial, and the result will be the sum of all the coefficients. Is that right? Let me verify. For example, if I have a simple polynomial like 2x + 3, the coefficients are 2 and 3. Their sum is 5. If I substitute x = 1, I get 2*1 + 3 = 5. Yeah, that works. So substituting x = 1 gives the sum of the coefficients. That's a useful trick. So instead of expanding the entire expression, which might take some time and could lead to errors, maybe I can just plug in x = 1 into the original expression and calculate the result. That should give me the sum of all the coefficients in the expanded form. Let me try that. Substituting x = 1 into (x³ + 2x + 1)(3x² + 4):First part: (1)³ + 2*(1) + 1 = 1 + 2 + 1 = 4.Second part: 3*(1)² + 4 = 3 + 4 = 7.Multiply them together: 4 * 7 = 28.So, the sum of the coefficients should be 28. Wait, but let me make sure that this method is applicable here. Since substitution works for any polynomial, even if it's a product of two polynomials, right? Because when you multiply two polynomials, the resulting polynomial's coefficients are combinations of the original coefficients. So substituting x = 1 into the product is the same as substituting x = 1 into each polynomial and then multiplying the results. Therefore, the sum of the coefficients of the product is indeed the product of the sums of the coefficients of each factor. That makes sense. But just to be thorough, maybe I should also expand the expression manually and add up the coefficients to check if I get the same answer. Let's try that. First, let's expand (x³ + 2x + 1)(3x² + 4). Multiply each term in the first polynomial by each term in the second polynomial:1. x³ * 3x² = 3x⁵2. x³ * 4 = 4x³3. 2x * 3x² = 6x³4. 2x * 4 = 8x5. 1 * 3x² = 3x²6. 1 * 4 = 4Now, combine all these terms:3x⁵ + 4x³ + 6x³ + 8x + 3x² + 4.Combine like terms:- 3x⁵- (4x³ + 6x³) = 10x³- 3x²- 8x- 4So the expanded polynomial is: 3x⁵ + 10x³ + 3x² + 8x + 4.Now, let's list the coefficients:- 3 (for x⁵)- 10 (for x³)- 3 (for x²)- 8 (for x)- 4 (constant term)Summing them up: 3 + 10 + 3 + 8 + 4.Calculating step by step:3 + 10 = 1313 + 3 = 1616 + 8 = 2424 + 4 = 28.Wow, that's the same result as before. So both methods give me 28. That confirms that the shortcut works here. Therefore, the sum of the coefficients of all terms in the expansion is 28. But just to make sure there are no mistakes in the expansion, let me check each multiplication step again:Original polynomials: (x³ + 2x + 1) and (3x² + 4)First term: x³ * 3x² = 3x^(3+2) = 3x⁵ ✔️x³ * 4 = 4x³ ✔️2x * 3x² = 6x^(1+2) = 6x³ ✔️2x * 4 = 8x ✔️1 * 3x² = 3x² ✔️1 * 4 = 4 ✔️Combining like terms: 3x⁵, 4x³ +6x³=10x³, 3x², 8x, 4. Correct. ✔️Coefficients: 3, 10, 3, 8, 4. Adding them gives 28. All steps check out. Alternatively, maybe there's another way to see why substituting x=1 works. Each coefficient in the expansion contributes to the sum when x=1 because 1 raised to any power is 1. Therefore, each term a_nx^n becomes a_n*1^n = a_n. Hence, the entire polynomial evaluated at x=1 is the sum of its coefficients. So regardless of how complicated the polynomial is, whether it's a product, sum, difference, etc., substituting x=1 gives the sum of coefficients. That's a fundamental property of polynomials. Therefore, whether we do the substitution first or expand first, we get the same result. Just to challenge myself, what if the question had asked for the sum of the absolute values of the coefficients? Then substituting x=1 wouldn't work, right? Because negative coefficients would become positive when taking absolute values, but substituting x=1 or any other number wouldn't account for that. So in that case, we would have to expand the polynomial and take absolute values before summing. But in this problem, all coefficients in the expanded form are positive. Let me check:Looking at the expanded polynomial: 3x⁵ + 10x³ + 3x² + 8x + 4. All coefficients are positive (3,10,3,8,4). So even if we took absolute values, the sum would be the same. But in the original factors, the coefficients were positive as well. So maybe in this specific case, even if the question had asked for absolute values, the answer would still be 28. But the problem doesn't specify absolute values, so we just take the coefficients as they are, which are all positive here. But the key takeaway is that for the standard sum of coefficients, substituting x=1 is a reliable method. So in exams or timed settings, that's a great shortcut. Another thought: If the problem had included negative coefficients, would that affect the sum? For example, if there was a term like -2x², the coefficient would be -2, and it would subtract from the total sum. But in our problem, all coefficients are positive, so no issues. But regardless, the method remains the same. Substitute x=1 into the expression, compute the value, and that's the sum of coefficients. So even if the expression had negative coefficients, substitution would handle it by subtracting accordingly. For example, (x - 1)(x + 1) = x² - 1. Substituting x=1: (1 -1)(1 +1) = 0*2 = 0. The sum of coefficients is 1 (for x²) + 0 (for x) + (-1) (constant term) = 0. So that works. Therefore, the method is solid. So in conclusion, whether through substitution or expansion, the sum of the coefficients in this problem is 28. I’m confident that’s the correct answer.Final Answerboxed{28}</think>To determine the sum of the coefficients of all terms in the expansion of ((x^3 + 2x + 1)(3x^2 + 4)), we can use the shortcut method of substituting (x = 1) into the polynomial expression. This method works because substituting (x = 1) into a polynomial results in the sum of all its coefficients.First, substitute (x = 1) into the given expression:[(1^3 + 2 cdot 1 + 1)(3 cdot 1^2 + 4)]Calculate each part separately:[(1 + 2 + 1)(3 + 4) = (4)(7) = 28]To verify, we can expand the polynomial and sum the coefficients. Expanding ((x^3 + 2x + 1)(3x^2 + 4)):1. (x^3 cdot 3x^2 = 3x^5)2. (x^3 cdot 4 = 4x^3)3. (2x cdot 3x^2 = 6x^3)4. (2x cdot 4 = 8x)5. (1 cdot 3x^2 = 3x^2)6. (1 cdot 4 = 4)Combining like terms:[3x^5 + (4x^3 + 6x^3) + 3x^2 + 8x + 4 = 3x^5 + 10x^3 + 3x^2 + 8x + 4]The coefficients are 3, 10, 3, 8, and 4. Summing these coefficients:[3 + 10 + 3 + 8 + 4 = 28]Both methods confirm that the sum of the coefficients is (boxed{28}).]

🔑:Upon expanding the given expression (x^{3}+2x+1)(3x^{2}+4), we obtain a polynomial containing the variable x.To find the sum of the coefficients of all terms in the expanded polynomial, we can substitute x=1 into the expression:(1^{3}+2(1)+1)(3(1)^{2}+4) = (1+2+1)(3+4) = 28.Therefore, the sum of the coefficients of all terms in the expanded polynomial is boxed{28}.By setting x=1, we can easily compute the sum of the coefficients of all terms in the expanded polynomial. This problem focuses on applying the concept of finding the sum of coefficients in an expanded polynomial. To solve this problem effectively, utilize the substitution x=1 to simplify the calculation process. This is a basic-level problem.

🧠:[<think>Okay, let me try to figure this out. The problem is about an odd function f(x) defined on all real numbers. It satisfies f(x + 1) = f(1 - x), and when x is between 0 and 1, inclusive, f(x) is log base 2 of (x + 1). We need to find f(31). The answer choices are A: 0, B:1, C:-1, D:2. Alright, let's break this down step by step.First, let's recall what an odd function is. An odd function satisfies the property f(-x) = -f(x) for all x in its domain. So, if we know the value of the function at some point x, we know it at -x with the opposite sign. That might come in handy here.Next, the function is defined on all real numbers, but we only know its explicit form on the interval [0,1]. For x in [0,1], f(x) = log₂(x + 1). So, if we can express 31 in terms that relate to this interval, using the given functional equations, we can compute f(31).The other important property given is f(x + 1) = f(1 - x). Let me parse this. If I let y = x + 1, then this equation becomes f(y) = f(2 - y). So, replacing y with x, we can say f(x) = f(2 - x) for all x. Wait, that seems like a symmetry around x = 1. Because if you reflect x over the line x = 1, you get 2 - x. So, the function is symmetric about x = 1. That's interesting. So, this is a kind of reflection symmetry.So, the function is both odd and has this symmetry about x = 1. Let's see how these two properties interact. Let's see:Since f is odd, f(-x) = -f(x). And it's symmetric about x = 1, so f(x) = f(2 - x). Let me try to see if these two can be combined.For example, let's pick some x and see what these properties imply. Suppose x = 1 + a, then f(1 + a) = f(2 - (1 + a)) = f(1 - a). So, f(1 + a) = f(1 - a). Which is similar to the original given equation f(x + 1) = f(1 - x). So, maybe this is the same as saying that f is symmetric around x = 1. So, every point a units to the right of 1 is the same as a units to the left of 1. So, the graph is symmetric around the line x = 1.But also, since the function is odd, it's symmetric about the origin. So, combining these two symmetries might lead to some periodicity? Let's check.Let me see. Let's try to see if f(x + 4) = f(x). If there's a period of 4, then f(31) would be the same as f(31 mod 4). Let's compute 31 divided by 4: 4*7=28, 31-28=3, so 31 mod 4 is 3. So, f(31) would be f(3). Then maybe f(3) can be related back to the interval [0,1] using the symmetries.But wait, how can we see if the function is periodic? Let's try to find the period. Let's see:Since f is symmetric about x = 1, f(x) = f(2 - x). Also, since f is odd, f(-x) = -f(x).Let's combine these. Let's replace x with 2 - x in the odd function property. So, f(-(2 - x)) = -f(2 - x). But f(2 - x) is f(x), so f(-2 + x) = -f(x). Hmm. Also, since the function is odd, f(-2 + x) = -f(2 - x). Wait, maybe that's not directly helpful. Let me try plugging specific values.Let's compute f(31). Let's try to relate 31 to a number closer to 0 using the symmetries.First, using the symmetry about x = 1: f(x) = f(2 - x). So, f(31) = f(2 - 31) = f(-29). Now, since f is odd, f(-29) = -f(29). So, f(31) = -f(29).Now, let's apply the same symmetry to f(29): f(29) = f(2 - 29) = f(-27). Then, f(-27) = -f(27). Thus, f(29) = -f(27), so f(31) = -(-f(27)) = f(27).Continuing this pattern:f(31) = f(27) = f(23) = ... etc. Each time we subtract 4. Let's see how many steps this takes. 31, 27, 23, 19, 15, 11, 7, 3. So, each step subtracts 4, and after 7 steps (since (31 - 3)/4 = 7), we get to f(3). Then, f(3) = f(2 - 3) = f(-1) = -f(1), because f is odd. Therefore, f(3) = -f(1). Then, f(31) = f(3) = -f(1). So, now we need to compute f(1).But wait, when x is in [0,1], f(x) = log₂(x + 1). So, x = 1 is in [0,1], so f(1) = log₂(1 + 1) = log₂(2) = 1. Therefore, f(1) = 1. Hence, f(3) = -1, so f(31) = -1. Therefore, the answer is C: -1.Wait, but let me verify this step-by-step again to make sure.Starting with f(31). Using the symmetry f(x) = f(2 - x):f(31) = f(2 - 31) = f(-29). Then, since f is odd, f(-29) = -f(29), so f(31) = -f(29).Now, f(29) = f(2 - 29) = f(-27) = -f(27). Therefore, f(31) = -(-f(27)) = f(27).Similarly, f(27) = f(2 - 27) = f(-25) = -f(25), so f(27) = -f(25), hence f(31) = -f(25).Wait, wait a second, this seems contradictory. Wait, maybe I made a mistake here.Wait, starting over:f(31) = f(2 - 31) = f(-29) = -f(29).Then f(29) = f(2 - 29) = f(-27) = -f(27), so f(31) = -(-f(27)) = f(27).Similarly, f(27) = f(2 - 27) = f(-25) = -f(25), so f(27) = -f(25), hence f(31) = -f(25).Wait, so then f(31) = f(27) = -f(25). Hmm, so the sign alternates each time?Wait, let me check with n being an integer. Let's denote f(4k + 3). Let's see.Wait, 31 divided by 4 is 7*4 + 3, so 31 = 4*7 + 3. Similarly, 27 = 4*6 + 3, 23 = 4*5 + 3, etc., down to 3 = 4*0 + 3. So, each time we subtract 4, we're staying in the form 4k + 3.So, let's see: Starting with 31.f(31) = f(2 - 31) = f(-29) = -f(29).f(29) = f(2 - 29) = f(-27) = -f(27).Therefore, f(31) = -(-f(27)) = f(27).Similarly, f(27) = f(-25) = -f(25).So, f(31) = f(27) = -f(25).Then, f(25) = f(-23) = -f(23), so f(31) = -(-f(23)) = f(23).Continuing this pattern, we alternate between negative and positive each time we subtract 4. Wait, but is that the case?Wait, let's do it step by step:1. f(31) = f(2 - 31) = f(-29) = -f(29) [since f is odd]2. f(29) = f(2 - 29) = f(-27) = -f(27)3. Therefore, f(31) = -(-f(27)) = f(27)4. f(27) = f(2 - 27) = f(-25) = -f(25)5. Therefore, f(31) = -f(25)6. f(25) = f(2 - 25) = f(-23) = -f(23)7. Therefore, f(31) = -(-f(23)) = f(23)8. f(23) = f(2 - 23) = f(-21) = -f(21)9. Therefore, f(31) = -f(21)10. f(21) = f(2 - 21) = f(-19) = -f(19)11. Therefore, f(31) = -(-f(19)) = f(19)12. Continuing this way, each time we subtract 4, we alternate between a negative and positive.So, for each subtraction of 4, starting from 31:31: f(31) = -f(29)29: f(29) = -f(27) => f(31) = f(27)27: f(27) = -f(25) => f(31) = -f(25)25: f(25) = -f(23) => f(31) = f(23)23: f(23) = -f(21) => f(31) = -f(21)21: f(21) = -f(19) => f(31) = f(19)19: f(19) = -f(17) => f(31) = -f(17)17: f(17) = -f(15) => f(31) = f(15)15: f(15) = -f(13) => f(31) = -f(13)13: f(13) = -f(11) => f(31) = f(11)11: f(11) = -f(9) => f(31) = -f(9)9: f(9) = -f(7) => f(31) = f(7)7: f(7) = -f(5) => f(31) = -f(5)5: f(5) = -f(3) => f(31) = f(3)3: f(3) = f(2 - 3) = f(-1) = -f(1) [since f is odd]Therefore, f(31) = -f(1)But f(1) is given for x in [0,1], so f(1) = log₂(1 + 1) = log₂(2) = 1Hence, f(31) = -1So, the answer should be C: -1.But let me check once more. Alternatively, perhaps there's a pattern here. Since each time we subtract 4, the sign alternates. So, starting from 31, how many steps of subtracting 4 do we take to get down to 3?From 31 to 3: 31 - 4*7 = 31 - 28 = 3. So, 7 steps. Each step involves two operations: applying the symmetry f(x) = f(2 - x) and then using the odd property f(-x) = -f(x). Each step reduces the argument by 4 and multiplies by -1 once. Wait, but in each step, we have two sign flips?Wait, let's think in terms of transformations. Let's define T(x) = 2 - x. Then, since f(T(x)) = f(x), we can say that applying T reflects over x=1. Similarly, since f is odd, reflecting over x=0 gives -f(-x). So, combining these reflections.Alternatively, the combination of the two symmetries (oddness and reflection over x=1) might lead to a translational symmetry. Let's see:Suppose we take x and apply the reflection over x=1 first: f(x) = f(2 - x). Then, since f is odd, f(2 - x) = -f(x - 2). So, combining these, f(x) = -f(x - 2). Therefore, f(x) = -f(x - 2). If we apply this again, f(x - 2) = -f(x - 4). Therefore, f(x) = -(-f(x - 4)) = f(x - 4). Hence, f(x) = f(x - 4). Therefore, the function is periodic with period 4.Ah! So, combining the two symmetries, f(x) = -f(x - 2), and then applying that again gives f(x) = f(x - 4). Therefore, the function is periodic with period 4. Therefore, f(31) = f(31 mod 4). 31 divided by 4 is 7*4=28, remainder 3. So, 31 mod 4 is 3. Therefore, f(31) = f(3).Then, to compute f(3), we can use the original symmetries. Since f(3) = f(2 - 3) = f(-1). And since f is odd, f(-1) = -f(1). Therefore, f(3) = -f(1). Then, f(1) is in the interval [0,1], so f(1) = log₂(1 + 1) = 1. Hence, f(3) = -1, so f(31) = -1.Therefore, the answer is C: -1. This matches our previous conclusion.So, this function is periodic with period 4. Let me verify this conclusion once more. Let's take an arbitrary x. Using the relation f(x) = -f(x - 2), which came from combining f(x) = f(2 - x) and f being odd. Then, applying the same relation to f(x - 2): f(x - 2) = -f(x - 4). Hence, substituting back into f(x) = -f(x - 2), we have f(x) = -(-f(x - 4)) = f(x - 4). Therefore, period 4.Therefore, f(x + 4) = f(x) for all x. Hence, f(31) = f(31 - 4*7) = f(3). Then, f(3) = -f(1) = -1. Therefore, correct.Thus, the answer is C: -1.Final Answerboxed{C}</think>Given that the odd function ( f(x) ) defined on (mathbb{R}) satisfies ( f(x+1) = f(1-x) ), and when ( x in [0,1] ), ( f(x) = log_{2}(x+1) ), we need to determine the value of ( f(31) ).1. Symmetry Property: The function ( f(x) ) satisfies ( f(x) = f(2 - x) ) for all ( x ). This indicates symmetry about ( x = 1 ).2. Odd Function Property: The function ( f(x) ) is odd, meaning ( f(-x) = -f(x) ).By combining these properties, we derive the periodicity of the function:- Using the symmetry property, ( f(x) = f(2 - x) ).- Using the odd function property, ( f(2 - x) = -f(x - 2) ).- Therefore, ( f(x) = -f(x - 2) ).- Applying this recursively, we find ( f(x) = -f(x - 2) ) and ( f(x - 2) = -f(x - 4) ), leading to ( f(x) = f(x - 4) ). Hence, ( f(x) ) is periodic with period 4.To find ( f(31) ):- Calculate ( 31 mod 4 ), which gives 3. Thus, ( f(31) = f(3) ).- Using the symmetry property, ( f(3) = f(2 - 3) = f(-1) ).- Using the odd function property, ( f(-1) = -f(1) ).- For ( x = 1 ) in the interval ([0, 1] ), ( f(1) = log_{2}(1 + 1) = 1 ).Therefore, ( f(3) = -f(1) = -1 ).The value of ( f(31) ) is (boxed{C}).,<think>Okay, let's see. The problem is about an odd function f(x) defined on all real numbers. It satisfies the condition f(x + 1) = f(1 - x). And when x is in the interval [0, 1], f(x) is given as log base 2 of (x + 1). We need to find f(31) and the options are A:0, B:1, C:-1, D:2.First, since f is an odd function, that means f(-x) = -f(x) for all x. That's the definition of an odd function. So whatever f(31) is, f(-31) should be -f(31). But since we need f(31), maybe we can find some symmetry or periodicity using the given condition f(x + 1) = f(1 - x).Let me try to parse that condition. For any real number x, if I shift x by 1 to the right, then the function value at x + 1 is the same as the function value at 1 - x. Let me see if I can find a pattern or maybe a period here.Let's make a substitution. Let y = x + 1. Then the equation becomes f(y) = f(1 - (y - 1)) = f(2 - y). So, f(y) = f(2 - y). That looks like a reflection around y = 1. So the function is symmetric about the line x = 1. So for any point y, its mirror image across x = 1 is the same function value. That's an interesting symmetry.But the function is also odd, which is a symmetry about the origin. Hmm. So combining these two symmetries, maybe we can find some periodicity?Wait, odd function means symmetric about the origin, and the other condition is symmetric about x = 1. Let's see if these two can lead to periodicity.Suppose we have f(y) = f(2 - y) from the substitution. Also, since f is odd, f(-y) = -f(y). Let's see if we can relate these.Let me try to apply both conditions. Let's take some y and see:From f(y) = f(2 - y) and f(-y) = -f(y). Let's substitute y with 2 - y in the oddness condition.Wait, f(- (2 - y)) = -f(2 - y). But f(2 - y) = f(y), so f(- (2 - y)) = -f(y). Let's compute that left side:f(-2 + y) = -f(y). So f(y - 2) = -f(y). That's an interesting relation. So shifting y by -2 gives the negative of the function value at y. Let's write that down: f(y - 2) = -f(y).So if I replace y with y + 2, then f(y + 2 - 2) = f(y) = -f(y + 2). Therefore, f(y) = -f(y + 2). So this gives a recurrence relation: f(y + 2) = -f(y). So every time we shift y by 2, the function flips sign. Then shifting again by 2, f(y + 4) = -f(y + 2) = -(-f(y)) = f(y). Therefore, the function is periodic with period 4. Because shifting by 4 brings us back to the original value. So periodicity is 4.So f is periodic with period 4. That means f(y + 4) = f(y) for all y. Therefore, to compute f(31), we can subtract multiples of 4 until we get within the fundamental period.Compute 31 divided by 4: 4*7 = 28, so 31 - 28 = 3. Therefore, 31 ≡ 3 mod 4. So f(31) = f(3). Then we need to compute f(3).But to compute f(3), let's use the recurrence relation. Since f(y + 2) = -f(y), then f(3) = f(1 + 2) = -f(1). So f(3) = -f(1). Now, since 1 is in the interval [0, 1], we can compute f(1) directly from the given expression: f(1) = log₂(1 + 1) = log₂(2) = 1. Therefore, f(3) = -1.Therefore, f(31) = f(3) = -1. So the answer should be C: -1.Wait, let me verify this step by step again to make sure.First, confirm that f is odd: f(-x) = -f(x). Then the given condition f(x + 1) = f(1 - x). So substituting variables as above, we found that f(y) = f(2 - y), symmetry about x = 1. Then combining with oddness, we found f(y - 2) = -f(y). Then replacing y with y + 2, f(y) = -f(y + 2). Then f(y + 2) = -f(y). So f(y + 4) = f(y). So period 4.Therefore, f(31) = f(31 - 4*7) = f(31 - 28) = f(3). Then f(3) = f(1 + 2) = -f(1). Since 1 is in [0,1], f(1) = log₂(1 + 1) = 1. Hence, f(3) = -1. So f(31) = -1. So the answer is C.Alternatively, maybe I can check with another approach. Let's try to extend the function beyond [0,1].Given that f(x) is defined on [0,1] as log₂(x + 1). Since f is odd, for x in [-1,0), f(x) = -f(-x) = -log₂(-x + 1). Because if x is in [-1,0), then -x is in (0,1], so f(-x) is log₂(-x + 1).So in the interval [-1,1], we have f(x) defined as log₂(x + 1) for x >=0 and -log₂(-x + 1) for x <0.But the function also satisfies f(x + 1) = f(1 - x). Let's see if we can use that to extend it beyond [-1,1].Let me take x in [0,1]. Then x + 1 is in [1,2]. So for x in [0,1], f(x + 1) = f(1 - x). But x is in [0,1], so 1 - x is in [0,1]. Therefore, f(1 - x) = log₂(1 - x + 1) = log₂(2 - x). Therefore, f(x + 1) = log₂(2 - x) for x in [0,1]. Therefore, for y in [1,2], f(y) = log₂(2 - (y - 1)) = log₂(3 - y).Wait, that's an interesting result. So for y in [1,2], f(y) = log₂(3 - y).But also, since f is odd, we can find f for negative values. Let's see.But maybe we can build the function over intervals step by step.First interval: [-1,1], we have f defined as above.Then using the relation f(y + 1) = f(1 - y). Wait, actually, we had the original condition f(x + 1) = f(1 - x). Let me check that again.Wait, the original condition is f(x + 1) = f(1 - x) for all x. So for any x, shifting x by 1 gives a value equal to the reflection over x=1. So for example, if x = 0, f(1) = f(1 - 0) = f(1), which is trivial. If x = 1, f(2) = f(0). Since f(0) = log₂(0 + 1) = log₂(1) = 0. So f(2) = 0. But from the periodicity, we found f(2) = -f(0) = -0 = 0, which matches.Similarly, if x = 2, f(3) = f(1 - 2) = f(-1). But f(-1) = -f(1) = -1. So f(3) = -1. Then f(3) = -1. Which matches our previous result.So f(3) = -1, so f(31) = f(3) = -1.Alternatively, since the period is 4, f(31) = f(31 mod 4) = f(3). Then f(3) = -f(1) = -1. So yes, answer is C.Wait, but perhaps to verify even more steps.Let me try to express the function in different intervals.Given that f is odd and satisfies f(x + 1) = f(1 - x). Let's use these two properties to define f over all real numbers.First, for x in [0,1], f(x) = log₂(x + 1).For x in [1,2], let’s set x = 1 + t where t ∈ [0,1]. Then f(1 + t) = f(1 - t) because of the condition f(x + 1) = f(1 - x). Wait, let me check that.Wait, the condition is f(x + 1) = f(1 - x). Let x = t, then f(t + 1) = f(1 - t). So if t ∈ [0,1], then t +1 ∈ [1,2], and 1 - t ∈ [0,1]. Therefore, for y ∈ [1,2], f(y) = f(2 - y). So f(y) = log₂((2 - y) + 1) = log₂(3 - y). Wait, but when y ∈ [1,2], 2 - y ∈ [0,1], so f(2 - y) = log₂( (2 - y) +1 ) = log₂(3 - y). Therefore, f(y) = log₂(3 - y) for y ∈ [1,2].Similarly, for x in [2,3], let y = x - 2. Then y ∈ [0,1]. Then using the periodicity we found earlier, f(x) = f(x - 4). Wait, no, but we need to use the recurrence relation f(y + 2) = -f(y). So f(x) = -f(x - 2). Therefore, for x ∈ [2,3], f(x) = -f(x - 2). Since x -2 ∈ [0,1], f(x -2) = log₂(x -2 +1 ) = log₂(x -1). Therefore, f(x) = -log₂(x -1) for x ∈ [2,3].Similarly, for x ∈ [3,4], we can apply f(x) = -f(x - 2). Since x - 2 ∈ [1,2], we know from before that f(x - 2) = log₂(3 - (x - 2)) = log₂(5 - x). Therefore, f(x) = -log₂(5 - x) for x ∈ [3,4].Then, using the periodicity of 4, f(x + 4) = f(x). Therefore, for any x beyond 4, we can subtract 4 until we get into the interval [0,4].But since 31 divided by 4 is 7*4 = 28, remainder 3. So 31 = 4*7 + 3, so f(31) = f(3). Then, in the interval [3,4], f(3) = -log₂(5 - 3) = -log₂(2) = -1. Therefore, f(31) = -1.So that confirms the previous result. Alternatively, using the oddness and the given conditions, we can directly compute f(31) = f(3) = -f(1) = -1.Another way to check: Let's list some values.f(0) = log₂(0 +1 ) = 0.f(1) = log₂(1 +1 ) = 1.By the condition f(x +1 ) = f(1 -x ), setting x = 0: f(1) = f(1), okay. x=1: f(2) = f(0) = 0. So f(2) =0.x=2: f(3) = f(-1). Since f is odd, f(-1) = -f(1) = -1. So f(3) = -1.x=3: f(4) = f(-2). f(-2) = -f(2) = -0 = 0. So f(4) =0.x=4: f(5) = f(-3). f(-3) = -f(3) = -(-1) =1. So f(5)=1.x=5: f(6) = f(-4)= -f(4)= -0=0. So f(6)=0.x=6: f(7)=f(-5)= -f(5)= -1. So f(7)=-1.And so on. So the pattern seems to be f(0)=0, f(1)=1, f(2)=0, f(3)=-1, f(4)=0, f(5)=1, f(6)=0, f(7)=-1, etc. So every 4 steps, it cycles. So indeed, f(4k) =0, f(4k+1)=1, f(4k+2)=0, f(4k+3)=-1.Since 31 divided by 4 is 7*4 +3, so 31=4*7+3, which is 4k+3, so f(31)= -1.Therefore, the answer is C: -1.Final Answerboxed{C}</think>Given that the odd function ( f(x) ) defined on ( mathbb{R} ) satisfies ( f(x+1) = f(1-x) ), and when ( x in [0,1] ), ( f(x) = log_{2}(x+1) ), we need to determine the value of ( f(31) ).1. Symmetry and Periodicity: - The function ( f(x) ) is odd, so ( f(-x) = -f(x) ). - The given condition ( f(x+1) = f(1-x) ) implies symmetry about ( x = 1 ). - By substituting ( y = x + 1 ), we get ( f(y) = f(2 - y) ). - Combining the odd function property and symmetry, we derive ( f(y - 2) = -f(y) ), leading to the periodicity relation ( f(y + 4) = f(y) ).2. Periodicity Analysis: - The function ( f(x) ) is periodic with period 4. - To find ( f(31) ), we reduce 31 modulo 4: ( 31 equiv 3 mod 4 ). Thus, ( f(31) = f(3) ).3. Evaluating ( f(3) ): - Using the periodicity and symmetry, ( f(3) = f(1 + 2) = -f(1) ). - For ( x = 1 ), ( f(1) = log_{2}(1 + 1) = 1 ). - Therefore, ( f(3) = -f(1) = -1 ).Thus, the value of ( f(31) ) is (boxed{C}).]

🔑:Since the odd function f(x) satisfies f(x+1)=f(1-x),We have f(x+1)=f(1-x)=-f(x-1), which means f(x+2)=-f(x),Thus, f(x+4)=-f(x+2)=f(x),This indicates that function f(x) has a period of 4.Given that when x in [0,1], f(x) = log_{2}(x+1),We can deduce that f(31)=f(32-1)=f(-1)=-f(1)=-log_{2}2=boxed{-1},Therefore, the answer is: boxed{text{C}}.This problem primarily requires calculating the function value. By determining the periodicity of the function from the given conditions and utilizing the function's odd symmetry and periodicity for conversion, we can solve this problem.