🧠:[<think>Okay, so I need to figure out which quadrant the complex number (frac{z_1}{z_2}) is in, where (z_1 = 1 - 2i) and (z_2 = 2 + 3i). Let me start by recalling how to divide complex numbers. From what I remember, to divide complex numbers, you multiply the numerator and denominator by the conjugate of the denominator. The conjugate of a complex number (a + bi) is (a - bi). This helps eliminate the imaginary part in the denominator. So, applying that here, the conjugate of (z_2 = 2 + 3i) should be (2 - 3i). Let me write that out step by step. First, write down (z_1/z_2):[frac{1 - 2i}{2 + 3i}]Now, multiply numerator and denominator by the conjugate of the denominator, which is (2 - 3i):[frac{(1 - 2i)(2 - 3i)}{(2 + 3i)(2 - 3i)}]Okay, now I need to perform the multiplication in both the numerator and the denominator. Let me start with the denominator since that might be simpler. Denominator:[(2 + 3i)(2 - 3i)]Using the formula ((a + b)(a - b) = a^2 - b^2), this becomes:[(2)^2 - (3i)^2 = 4 - 9i^2]But since (i^2 = -1), this simplifies to:[4 - 9(-1) = 4 + 9 = 13]So the denominator is 13. Good, that's a real number, which is what we wanted.Now the numerator:[(1 - 2i)(2 - 3i)]I need to expand this using the distributive property (FOIL method). Let's do that:First: (1 times 2 = 2)Outer: (1 times (-3i) = -3i)Inner: (-2i times 2 = -4i)Last: (-2i times (-3i) = 6i^2)Combine all these terms:[2 - 3i - 4i + 6i^2]Combine like terms:The real parts: 2 and (6i^2). Wait, (i^2 = -1), so (6i^2 = -6). So the real part is (2 - 6 = -4).The imaginary parts: (-3i - 4i = -7i)So the numerator simplifies to:[-4 - 7i]Therefore, putting numerator over denominator:[frac{-4 - 7i}{13} = -frac{4}{13} - frac{7}{13}i]So the result of the division is the complex number (-frac{4}{13} - frac{7}{13}i). Now, to determine the quadrant in the complex plane, we look at the real part (x-axis) and the imaginary part (y-axis). In this case, the real part is (-frac{4}{13}) which is negative, and the imaginary part is (-frac{7}{13}) which is also negative. In the complex plane, quadrants are determined similar to the Cartesian plane:- Quadrant I: Positive real, positive imaginary- Quadrant II: Negative real, positive imaginary- Quadrant III: Negative real, negative imaginary- Quadrant IV: Positive real, negative imaginarySince both the real and imaginary parts are negative here, the point (-frac{4}{13} - frac{7}{13}i) is in Quadrant III. Wait, let me double-check my calculations to make sure I didn't make a mistake. First, denominator calculation:[(2 + 3i)(2 - 3i) = 4 - 6i + 6i - 9i^2 = 4 - 9(-1) = 13. quad text{That's correct.}]Numerator:[(1)(2) + (1)(-3i) + (-2i)(2) + (-2i)(-3i)][= 2 - 3i -4i + 6i^2][= 2 -7i +6(-1)][= 2 -7i -6 = -4 -7i quad text{Right.}]So the final result is (-frac{4}{13} - frac{7}{13}i), which is indeed in Quadrant III. But hold on, looking at the answer choices: A. Quadrant I, B. Quadrant II, C. Quadrant III, D. Quadrant IV. So according to this, the answer should be C. Quadrant III. But wait, let me confirm once again. Maybe I miscalculated the signs? Let's check:Original numerator:First term: 1*2 = 2Second term: 1*(-3i) = -3iThird term: (-2i)*2 = -4iFourth term: (-2i)*(-3i) = 6i² = 6*(-1) = -6So adding up real parts: 2 + (-6) = -4Imaginary parts: (-3i) + (-4i) = -7iSo numerator is -4 -7i. Divided by 13, gives real part -4/13, imaginary part -7/13. So both coordinates negative. So yes, Quadrant III. So answer C.Wait, but the answer options are A through D, with Quadrant IV being D. Hmm. Wait, is Quadrant III the one where both real and imaginary parts are negative? Yes. In the complex plane, Quadrant I is top right, Quadrant II is top left, Quadrant III is bottom left, and Quadrant IV is bottom right. So negative real and negative imaginary is Quadrant III, so answer C.But let me check once more if I didn't miscalculate the multiplication. Maybe when I did the numerator, I messed up the signs. Let me recast the problem another way. Alternatively, perhaps using another method.Alternatively, you can represent the complex numbers in polar form and divide them, but that might be more complicated. But perhaps converting the final result into coordinates is straightforward. Let me see.Alternatively, maybe I miscalculated the original division. Let me try another approach.Alternatively, compute (frac{z_1}{z_2}) by using (z_1 = 1 -2i) and (z_2 = 2 + 3i). Let me write it as:[frac{1 - 2i}{2 + 3i}]Multiply numerator and denominator by the conjugate of the denominator (2 - 3i):Numerator: ((1 - 2i)(2 - 3i))Denominator: ((2 + 3i)(2 - 3i))As before, denominator is 13. Numerator: 1*2 + 1*(-3i) + (-2i)*2 + (-2i)*(-3i) = 2 -3i -4i +6i². Then, 2 -7i -6 = -4 -7i. So same result.So conclusion is the same. So the point is (-4/13, -7/13). Both coordinates negative, so Quadrant III. Answer C.But wait, the options given are A. Quadrant I, B. Quadrant II, C. Quadrant III, D. Quadrant IV. So the correct answer is C. But wait, maybe I need to check again.Wait, let me think. Is there any chance that the real part is positive? If my numerator had been positive, then... but no. Wait, in the numerator, after expanding, we had 2 -3i -4i +6i². 6i² is 6*(-1)= -6. Then 2 -6 = -4, so real part -4. Then imaginary parts -3i -4i = -7i. So yes, real part -4/13, imaginary part -7/13.Alternatively, maybe when I multiplied (-2i)*(-3i), that term is 6i² which is -6. So correct. So yes, numerator is -4 -7i. So the complex number is (-4/13, -7/13). So plotting this in the complex plane, it's in the third quadrant. Hence, answer C. Quadrant III.But wait, let me check once more. Let me substitute the numbers. If I compute, say, -4/13 is approximately -0.3077, and -7/13 is approximately -0.5385. So both coordinates are negative. So in the complex plane, the x-axis (real part) is negative and y-axis (imaginary part) is negative, which is Quadrant III. Yes, that's correct.Therefore, the answer should be C. Quadrant III.Final Answerboxed{C}</think>To determine the quadrant in the complex plane that corresponds to the point represented by (frac{z_1}{z_2}) where (z_1 = 1 - 2i) and (z_2 = 2 + 3i), we start by dividing the complex numbers.First, we multiply the numerator and denominator by the conjugate of the denominator (z_2):[frac{1 - 2i}{2 + 3i} times frac{2 - 3i}{2 - 3i} = frac{(1 - 2i)(2 - 3i)}{(2 + 3i)(2 - 3i)}]Calculating the denominator:[(2 + 3i)(2 - 3i) = 2^2 - (3i)^2 = 4 - 9(-1) = 4 + 9 = 13]Expanding the numerator:[(1 - 2i)(2 - 3i) = 1 cdot 2 + 1 cdot (-3i) + (-2i) cdot 2 + (-2i) cdot (-3i)][= 2 - 3i - 4i + 6i^2][= 2 - 7i + 6(-1)][= 2 - 7i - 6 = -4 - 7i]Thus, the result of the division is:[frac{-4 - 7i}{13} = -frac{4}{13} - frac{7}{13}i]The real part is (-frac{4}{13}) and the imaginary part is (-frac{7}{13}). Both parts are negative, which places the point in Quadrant III.[boxed{C}],<think>Okay, so I need to figure out which quadrant the complex number (frac{z_1}{z_2}) lies in, where (z_1 = 1 - 2i) and (z_2 = 2 + 3i). Hmm, quadrants in the complex plane are similar to the Cartesian coordinate system, right? So, if a complex number is (a + bi), then the real part is (a) and the imaginary part is (b). The quadrants are determined by the signs of these parts: - Quadrant I: Positive real and positive imaginary.- Quadrant II: Negative real and positive imaginary.- Quadrant III: Negative real and negative imaginary.- Quadrant IV: Positive real and negative imaginary.So, my goal is to compute (frac{z_1}{z_2}) and check the signs of the real and imaginary parts. First, I remember that to divide complex numbers, you multiply the numerator and denominator by the conjugate of the denominator. The conjugate of (z_2 = 2 + 3i) is (2 - 3i). Let me write that down:[frac{z_1}{z_2} = frac{1 - 2i}{2 + 3i} times frac{2 - 3i}{2 - 3i}]Multiplying the numerators and denominators:First, let's compute the numerator: ((1 - 2i)(2 - 3i)). Let me use the distributive property (FOIL method):- First: (1 times 2 = 2)- Outer: (1 times (-3i) = -3i)- Inner: (-2i times 2 = -4i)- Last: (-2i times (-3i) = 6i^2)Combine these terms:(2 - 3i - 4i + 6i^2)Combine like terms:(2 - 7i + 6i^2)But remember that (i^2 = -1), so (6i^2 = -6). Substitute that in:(2 - 7i - 6 = (2 - 6) - 7i = -4 - 7i)So the numerator is (-4 - 7i).Now the denominator: ((2 + 3i)(2 - 3i)). This is a difference of squares:(2^2 - (3i)^2 = 4 - 9i^2)Again, (i^2 = -1), so this becomes:(4 - 9(-1) = 4 + 9 = 13)So the denominator is 13.Putting it all together:[frac{z_1}{z_2} = frac{-4 - 7i}{13} = -frac{4}{13} - frac{7}{13}i]So, the real part is (-frac{4}{13}) and the imaginary part is (-frac{7}{13}). Now, check the signs. Both the real part and the imaginary part are negative. That would place the complex number in Quadrant III, right? Because Quadrant III is where both real and imaginary parts are negative.Wait, but let me double-check my calculations to make sure I didn't make a mistake. Sometimes when multiplying complex numbers, it's easy to slip up with the signs.Starting again with the numerator:((1 - 2i)(2 - 3i))Multiply 1 by 2: 21 by -3i: -3i-2i by 2: -4i-2i by -3i: 6i^2 (which is -6)Adding them up: 2 - 3i -4i -6 = (2 -6) + (-3i -4i) = -4 -7i. That's the same as before. So numerator is correct.Denominator: (2 + 3i)(2 - 3i) = 4 -6i +6i -9i^2. Wait, hold on, is that correct? Wait, actually, (a + b)(a - b) = a² - b². So here, a = 2, b = 3i. Therefore, it's 2² - (3i)^2 = 4 - 9i². Since i² = -1, that's 4 - (-9) = 13. So denominator is 13. That's correct.So the division gives (frac{-4 -7i}{13}), so real part is -4/13 and imaginary part is -7/13. Both negative. Therefore, Quadrant III, which is option C.But let me check the answer options again:A. Quadrant IB. Quadrant IIC. Quadrant IIID. Quadrant IVSo the answer should be C.Wait, but maybe I made a mistake in the sign when distributing? Let me check the numerator again step by step.First term: 1 * 2 = 2Outer term: 1 * (-3i) = -3iInner term: -2i * 2 = -4iLast term: -2i * (-3i) = 6i²Yes, that's 6i², which is -6. Then adding all terms:2 -3i -4i -6 = (2 -6) + (-3i -4i) = -4 -7i. Correct.So no mistake there.So the real part is negative, imaginary part is negative. Therefore, Quadrant III. So the answer is C. Hmm, but wait, let me check again if the problem is written correctly.The problem says z1 = 1 - 2i and z2 = 2 + 3i. Yes, so z1 is in Quadrant IV (real positive, imaginary negative) and z2 is in Quadrant I (both positive). Then dividing them. Hmm, but the result is in Quadrant III. So that's the conclusion.Alternatively, maybe there is an alternative method. For example, converting to polar form and dividing the magnitudes and subtracting the angles.But since the problem is about quadrants, which is about the signs, maybe we can think about the angle. Let's see.But maybe this is overcomplicating. Alternatively, another way: compute the real and imaginary parts as I did, and check the signs. Since both are negative, Quadrant III.Therefore, the answer should be C.But wait, let me think again. Maybe I confused numerator and denominator? Wait, no. The conjugate is multiplied to both numerator and denominator. So the denominator becomes a real number, which is 13, and numerator is -4 -7i. So yes, the result is correct.Alternatively, if I compute the division in another way: (frac{1 - 2i}{2 + 3i}). Let me compute the real and imaginary parts step by step.If z = a + bi = (frac{1 - 2i}{2 + 3i}), then:Multiply numerator and denominator by the conjugate of denominator:(frac{(1 - 2i)(2 - 3i)}{(2 + 3i)(2 - 3i)})We already computed that as (frac{-4 -7i}{13}), which is correct.Alternatively, using the formula for division:If ( z_1 = a + bi ), ( z_2 = c + di ), then:( frac{z_1}{z_2} = frac{(a + bi)(c - di)}{c^2 + d^2} )So here, a =1, b = -2, c=2, d=3.So numerator: (1)(2) + (1)(-3i) + (-2i)(2) + (-2i)(-3i) = 2 -3i -4i +6i²Wait, same as before.So real part is 2 + 6i² = 2 + 6(-1) = 2 -6 = -4.Imaginary part is (-3i -4i) = -7i.So indeed, real part is -4, imaginary part -7. So same result.Therefore, Quadrant III. So answer is C.But just to confirm, maybe plugging into another method.Alternatively, using complex plane angles.The angle of the quotient is the difference of the angles of z1 and z2. If the resulting angle is in Quadrant III, then that's where it is.But perhaps computing the angle is more complicated. Alternatively, compute the signs.But since z1 is in Quadrant IV (1, -2), so its angle is between 270 and 360 degrees. z2 is in Quadrant I (2, 3), angle between 0 and 90. So subtracting the angles: angle of z1/z2 is angle(z1) - angle(z2). If angle(z1) is, say, 315 degrees (as an example), angle(z2) is, say, 56.3 degrees (arctan(3/2)), then 315 - 56.3 = 258.7 degrees, which is in Quadrant IV? Wait, 180 to 270 is Quadrant III, 270 to 360 is Quadrant IV. 258.7 is between 180 and 270, so Quadrant III. Wait, wait, 258.7 is 180 + 78.7, so yes, Quadrant III.But if angle(z1) is 315, angle(z2) is 56.3, then 315 - 56.3 = 258.7, which is in Quadrant III. But in reality, z1's angle is arctan(-2/1) = arctan(-2), which is in Quadrant IV, so angle is 360 - arctan(2), which is approximately 360 - 63.43 = 296.57 degrees. z2's angle is arctan(3/2), which is approximately 56.31 degrees. Then the difference is 296.57 - 56.31 = 240.26 degrees. Which is between 180 and 270, so Quadrant III. So that matches our previous result. Hence, Quadrant III.Therefore, the answer is C. Quadrant III.Final Answerboxed{C}</think>To determine the quadrant in the complex plane that corresponds to the point represented by (frac{z_1}{z_2}) where (z_1 = 1 - 2i) and (z_2 = 2 + 3i), we need to perform the division of the complex numbers.First, we multiply the numerator and denominator by the conjugate of the denominator:[frac{z_1}{z_2} = frac{1 - 2i}{2 + 3i} times frac{2 - 3i}{2 - 3i}]Calculating the numerator:[(1 - 2i)(2 - 3i) = 1 cdot 2 + 1 cdot (-3i) + (-2i) cdot 2 + (-2i) cdot (-3i)][= 2 - 3i - 4i + 6i^2][= 2 - 7i + 6(-1)][= 2 - 7i - 6][= -4 - 7i]Calculating the denominator:[(2 + 3i)(2 - 3i) = 2^2 - (3i)^2][= 4 - 9i^2][= 4 - 9(-1)][= 4 + 9][= 13]Putting it all together:[frac{z_1}{z_2} = frac{-4 - 7i}{13} = -frac{4}{13} - frac{7}{13}i]The real part is (-frac{4}{13}) and the imaginary part is (-frac{7}{13}). Both parts are negative, which places the complex number in Quadrant III.Thus, the answer is (boxed{C}).]

🔑:To find the point corresponding to frac{z_1}{z_2} in the complex plane, we first calculate the value of the fraction.frac{z_1}{z_2} = frac{1-2i}{2+3i} = frac{(1-2i)(2-3i)}{(2+3i)(2-3i)} = frac{-4-7i}{13}Now, we can express this complex number in the form a+bi, where a=-frac{4}{13} and b=-frac{7}{13}. The point left(-frac{4}{13}, -frac{7}{13}right) in the complex plane lies in the third quadrant.Therefore, the answer is boxed{C}.This solution uses basic complex number operations and their geometric interpretation. The problem tests both reasoning and computational abilities and can be classified as a foundational question.

🧠:[<think>Okay, so I have this problem here: If (ab)/(a + b) = 3, (bc)/(b + c) = 6, and (ac)/(a + c) = 9, then find c/(ab). Hmm, let's see. I need to figure out the value of c divided by the product of a and b. First, let me write down the equations again to make sure I have them right:1. (ab)/(a + b) = 32. (bc)/(b + c) = 63. (ac)/(a + c) = 9And we need to find c/(ab). Hmm, okay. So these are three equations with three variables: a, b, c. Maybe I can solve for a, b, c first and then compute c/(ab). Let me try that approach. But solving three equations can be tricky. Let me see if I can manipulate these equations to express them in terms of reciprocals. Because each equation has a structure similar to something like (xy)/(x + y) = k, which might be easier if I take reciprocals. Let's consider the first equation: (ab)/(a + b) = 3. If I take the reciprocal of both sides, it becomes (a + b)/(ab) = 1/3. Which can be written as (1/a) + (1/b) = 1/3. Oh, that's clever. So, maybe if I let x = 1/a, y = 1/b, z = 1/c, then these equations can be transformed into linear equations. Let me try that substitution. Let x = 1/a, y = 1/b, z = 1/c. Then:From the first equation: (ab)/(a + b) = 3 → (1/(x y))/(1/x + 1/y) = 3? Wait, maybe not. Let me recheck. Wait, original equation is (ab)/(a + b) = 3. If I take reciprocal, (a + b)/ab = 1/3. Which is equal to 1/a + 1/b = 1/3. So that's x + y = 1/3. Similarly, the second equation: (bc)/(b + c) = 6. Reciprocal is (b + c)/bc = 1/6. So 1/b + 1/c = 1/6. Which in terms of y and z is y + z = 1/6. Third equation: (ac)/(a + c) = 9. Reciprocal is (a + c)/ac = 1/9. So 1/a + 1/c = 1/9. Which is x + z = 1/9. So now, after substitution, we have three linear equations:1. x + y = 1/32. y + z = 1/63. x + z = 1/9Ah, this seems manageable. So now, the problem is transformed into solving for x, y, z, then converting back to a, b, c. Then, once we have those, compute c/(ab). Let's proceed step by step.So equations:1. x + y = 1/32. y + z = 1/63. x + z = 1/9We need to solve for x, y, z. Let's subtract the third equation from the first equation: (x + y) - (x + z) = (1/3) - (1/9). So this gives y - z = 2/9. But from the second equation, we have y + z = 1/6. So now we have:y - z = 2/9y + z = 1/6If we add these two equations: (y - z) + (y + z) = 2/9 + 1/6 → 2y = (4/18 + 3/18) = 7/18 → y = 7/36.Then substituting back into y + z = 1/6: 7/36 + z = 6/36 → z = 6/36 - 7/36 = -1/36. Wait, z is negative? But z = 1/c. If z is negative, then c is negative. Is that possible? The problem didn't specify that a, b, c are positive. So maybe it's allowed. Let's proceed.Then from equation 1: x + y = 1/3, so x + 7/36 = 12/36 → x = 12/36 - 7/36 = 5/36.And from equation 3: x + z = 1/9, so 5/36 + z = 4/36 → z = 4/36 - 5/36 = -1/36. Yep, that matches. So z = -1/36.Therefore, we have:x = 5/36, so a = 1/x = 36/5.y = 7/36, so b = 1/y = 36/7.z = -1/36, so c = 1/z = -36.So a = 36/5, b = 36/7, c = -36.Now, we need to compute c/(ab). Let's compute ab first.ab = (36/5)*(36/7) = (36*36)/(5*7) = 1296/35.Then c/(ab) = (-36)/(1296/35) = (-36)*(35/1296).Simplify this. Let's compute 1296 divided by 36: 1296 ÷ 36 = 36. So 1296 = 36*36. Therefore:(-36)*(35/1296) = (-36)*(35)/(36*36) = (-35)/36.Therefore, c/(ab) = -35/36.Wait, but let me check the calculations again, because fractions can be tricky.First, a = 36/5, b = 36/7. Therefore, ab = (36/5)(36/7) = (36*36)/(5*7) = 1296/35. Correct.c is -36, so c/(ab) = (-36)/(1296/35) = (-36) * (35/1296). Now, 36 and 1296 have a common factor. 1296 divided by 36 is 36, as 36*36 = 1296. Therefore, 35/1296 divided by 36 is 35/(36*36) multiplied by 36. Wait, no. Wait: (-36) * (35/1296) = (-36 * 35)/1296. 36 is a factor of both numerator and denominator. 36/1296 = 1/36. Therefore, it's (-35)/36. So yes, -35/36. But the problem didn't specify whether it wants the answer as a positive fraction or if negative is acceptable. Since the problem didn't restrict the variables to be positive, and mathematically, the solution allows c to be negative, so the answer is -35/36. But let me double-check all steps to be sure. Let me verify the values of a, b, c.Given x = 1/a = 5/36 → a = 36/5. Correct.y = 1/b = 7/36 → b = 36/7. Correct.z = 1/c = -1/36 → c = -36. Correct.Now check if these values satisfy the original equations.First equation: (ab)/(a + b) = 3.Compute ab: (36/5)(36/7) = 1296/35.Compute a + b: 36/5 + 36/7 = (36*7 + 36*5)/35 = (252 + 180)/35 = 432/35.Therefore, (ab)/(a + b) = (1296/35)/(432/35) = 1296/432 = 3. Correct.Second equation: (bc)/(b + c) = 6.Compute bc: (36/7)(-36) = (-1296)/7.Compute b + c: 36/7 + (-36) = 36/7 - 252/7 = (-216)/7.Therefore, (bc)/(b + c) = (-1296/7)/(-216/7) = 1296/216 = 6. Correct.Third equation: (ac)/(a + c) = 9.Compute ac: (36/5)(-36) = (-1296)/5.Compute a + c: 36/5 + (-36) = 36/5 - 180/5 = (-144)/5.Therefore, (ac)/(a + c) = (-1296/5)/(-144/5) = 1296/144 = 9. Correct.All equations check out. So c/(ab) = -35/36.But let me check once more: c is -36, ab is 1296/35. Then c/(ab) is -36/(1296/35) = -36*(35/1296) = (-36*35)/1296. Divide numerator and denominator by 36: (-35)/36. Yep, correct. So the answer is -35/36.Wait, but the problem might expect a positive answer? Hmm, but based on the equations, c has to be negative here. Let me see. If all variables were positive, the third equation (ac)/(a + c) = 9 would require both a and c to be positive, so their product and sum positive. But the first equation gives positive a and b, so a is positive. But the third equation would then require c to be positive as well, but according to our solution, c is negative. Wait, so does this mean there is a mistake?Wait a minute, hold on. If a and c are both positive, then (ac)/(a + c) would be positive. But in our solution, c is negative. So that would make ac negative and a + c could be positive or negative. But according to the original problem, the left-hand sides of the equations are 3, 6, 9, which are positive. Therefore, the fractions (ab)/(a + b), (bc)/(b + c), (ac)/(a + c) must all be positive. Therefore, for each of these fractions to be positive, numerator and denominator must have the same sign. So, for (ab)/(a + b) = 3 > 0, so ab and a + b must be both positive or both negative. Similarly for the others.But in our solution, a is 36/5 (positive), b is 36/7 (positive), so ab is positive, and a + b is positive. So first equation is okay.Second equation: (bc)/(b + c) = 6. Here, b is positive, c is negative. Therefore, bc is negative, and b + c is 36/7 - 36. Let's compute that: 36/7 - 252/7 = (-216)/7, which is negative. So bc is negative, b + c is negative. Negative over negative is positive, which is 6. Correct.Third equation: (ac)/(a + c) = 9. a is positive, c is negative. So ac is negative. a + c is 36/5 - 36 = (36/5 - 180/5) = (-144)/5, which is negative. So negative divided by negative is positive, 9. Correct. Therefore, even though c is negative, the equations hold because the numerator and denominator are both negative, making the fraction positive. So our solution is valid. Therefore, c/(ab) is indeed -35/36.But let me check again the question: It says "then c/(ab) =". So they just want the value, which is negative. So unless there's a mistake in calculation, that's the answer. Alternatively, maybe there's another approach. Let's see. Since we need c/(ab), perhaps we can express c/(ab) in terms of the given equations without solving for a, b, c. Let me think.We have:From first equation: ab = 3(a + b)From second equation: bc = 6(b + c)From third equation: ac = 9(a + c)We need c/(ab). Since ab is 3(a + b), so c/(ab) = c/(3(a + b)) = (1/3)*(c/(a + b))But not sure if that helps. Alternatively, maybe express c in terms of a and b from the third equation? Let's see.From the third equation: ac = 9(a + c). Let's solve for c.ac = 9a + 9c → ac - 9c = 9a → c(a - 9) = 9a → c = 9a/(a - 9)So c is expressed in terms of a. Then, if we can express b in terms of a, we can substitute into ab = 3(a + b) and solve for a.Similarly, from the first equation: ab = 3(a + b) → ab - 3a - 3b = 0. Let's solve for b: b(a - 3) = 3a → b = 3a/(a - 3)So b = 3a/(a - 3)Similarly, from the second equation: bc = 6(b + c). Let's substitute c from above and b in terms of a.We have c = 9a/(a - 9), and b = 3a/(a - 3). Let's substitute into the second equation:[3a/(a - 3)] * [9a/(a - 9)] = 6[3a/(a - 3) + 9a/(a - 9)]Simplify left side: 27a² / [(a - 3)(a - 9)]Right side: 6[ (3a(a - 9) + 9a(a - 3)) / ((a - 3)(a - 9)) ) ] Let me compute the numerator inside the brackets on the right:3a(a - 9) + 9a(a - 3) = 3a² - 27a + 9a² - 27a = 12a² - 54aTherefore, the right side becomes 6*(12a² - 54a)/[(a - 3)(a - 9)] = (72a² - 324a)/[(a - 3)(a - 9)]So equate left and right sides:27a² / [(a - 3)(a - 9)] = (72a² - 324a)/[(a - 3)(a - 9)]Since denominators are the same, set numerators equal:27a² = 72a² - 324a → 0 = 45a² - 324a → 45a² - 324a = 0 → 9a(5a - 36) = 0Thus, a = 0 or a = 36/5But a = 0 would make the original equations undefined (division by zero in the reciprocal expressions), so a = 36/5.Therefore, a = 36/5. Then, b = 3a/(a - 3) = 3*(36/5)/(36/5 - 15/5) = (108/5)/(21/5) = 108/21 = 36/7. Which matches our previous result.Then c = 9a/(a - 9) = 9*(36/5)/(36/5 - 45/5) = (324/5)/(-9/5) = -324/9 = -36. Again, same as before.Therefore, this method also leads to a = 36/5, b = 36/7, c = -36. So c/(ab) = -36 / ( (36/5)(36/7) ) = -36 / (1296/35) = -36*(35/1296) = -35/36. Same answer.Therefore, this confirms the answer is -35/36.Alternatively, maybe we can use the reciprocals x, y, z again but in another way. Since we found x, y, z as 5/36, 7/36, -1/36, then c = 1/z = -36, and ab = (1/x)(1/y) = (1/(5/36))(1/(7/36)) = (36/5)(36/7) = 1296/35. Then c/(ab) = -36/(1296/35) = -35/36. So same result.So regardless of the method, the answer comes out to -35/36. Therefore, I think that's the correct answer.Final Answerboxed{-dfrac{35}{36}}</think>Given the equations:1. (frac{ab}{a + b} = 3)2. (frac{bc}{b + c} = 6)3. (frac{ac}{a + c} = 9)We need to find (frac{c}{ab}).First, we take the reciprocal of each equation and let (x = frac{1}{a}), (y = frac{1}{b}), and (z = frac{1}{c}). This transforms the equations into:1. (x + y = frac{1}{3})2. (y + z = frac{1}{6})3. (x + z = frac{1}{9})We solve these linear equations step by step:1. Subtracting the third equation from the first: ( (x + y) - (x + z) = frac{1}{3} - frac{1}{9} ) [ y - z = frac{2}{9} ]2. Adding the result to the second equation: ( (y - z) + (y + z) = frac{2}{9} + frac{1}{6} ) [ 2y = frac{7}{18} implies y = frac{7}{36} ]3. Substituting (y = frac{7}{36}) into the second equation: ( frac{7}{36} + z = frac{1}{6} ) [ z = frac{1}{6} - frac{7}{36} = -frac{1}{36} ]4. Substituting (y = frac{7}{36}) into the first equation: ( x + frac{7}{36} = frac{1}{3} ) [ x = frac{1}{3} - frac{7}{36} = frac{5}{36} ]Converting back to (a), (b), and (c):- (a = frac{1}{x} = frac{36}{5})- (b = frac{1}{y} = frac{36}{7})- (c = frac{1}{z} = -36)Calculating (frac{c}{ab}):- (ab = left(frac{36}{5}right) left(frac{36}{7}right) = frac{1296}{35})- (frac{c}{ab} = frac{-36}{frac{1296}{35}} = -36 times frac{35}{1296} = -frac{35}{36})Thus, the final answer is:[boxed{-dfrac{35}{36}}],<think>Okay, so I have this problem here: If (ab)/(a+b) = 3, (bc)/(b+c) = 6, and (ac)/(a+c) = 9, then find c/(ab). Hmm, let me think about how to approach this. First, the equations given are all fractions equal to some number. They look similar, so maybe there's a pattern here. Let me write them down again to see:1. (ab)/(a + b) = 32. (bc)/(b + c) = 63. (ac)/(a + c) = 9And the question is asking for c/(ab). Hmm. So maybe if I can find the values of a, b, c, then I can compute c/(ab) directly. But solving for a, b, c might be tricky. Let me see.Alternatively, maybe I can manipulate these equations to express a, b, c in terms of each other. Let's start with the first equation. If (ab)/(a + b) = 3, then taking reciprocals on both sides, (a + b)/ab = 1/3. Which can be written as 1/a + 1/b = 1/3. Wait, that's a good point. If I let x = 1/a, y = 1/b, z = 1/c, then maybe the equations become linear in x, y, z. That might be easier to solve. Let's try that substitution.Let me define:x = 1/ay = 1/bz = 1/cThen the first equation (ab)/(a + b) = 3 can be rewritten as:ab/(a + b) = 3Divide numerator and denominator by ab:1/( (1/a) + (1/b) ) = 3Which is 1/(x + y) = 3So that gives x + y = 1/3Similarly, the second equation (bc)/(b + c) = 6 becomes:1/( (1/b) + (1/c) ) = 6 => 1/(y + z) = 6 => y + z = 1/6Third equation (ac)/(a + c) = 9 becomes:1/( (1/a) + (1/c) ) = 9 => 1/(x + z) = 9 => x + z = 1/9Now, so we have a system of three equations:1. x + y = 1/32. y + z = 1/63. x + z = 1/9So now, this is a linear system. Let me write it down:Equation 1: x + y = 1/3Equation 2: y + z = 1/6Equation 3: x + z = 1/9Okay, let's solve for x, y, z. Let's subtract Equation 3 from Equation 1:(x + y) - (x + z) = (1/3) - (1/9)Simplify left side: x + y - x - z = y - zRight side: (3/9 - 1/9) = 2/9So, y - z = 2/9But from Equation 2, we have y + z = 1/6So now, we have:y - z = 2/9y + z = 1/6Let me add these two equations:(y - z) + (y + z) = 2/9 + 1/6Left side: 2yRight side: 2/9 + 1/6. Let's find a common denominator. 18. So 2/9 is 4/18, 1/6 is 3/18. So 4/18 + 3/18 = 7/18. Therefore, 2y = 7/18 => y = 7/36Now that we have y = 7/36, substitute back into Equation 1: x + y = 1/3 => x = 1/3 - y = 1/3 - 7/36. Convert 1/3 to 12/36. So 12/36 - 7/36 = 5/36. Therefore, x = 5/36.Similarly, from Equation 2: y + z = 1/6 => z = 1/6 - y = 1/6 - 7/36. Convert 1/6 to 6/36. So 6/36 - 7/36 = -1/36. Therefore, z = -1/36.Wait, z is -1/36? But z is 1/c. So if z is negative, that would mean c is negative. Hmm, is that possible? Let me check my calculations again.Equation 3: x + z = 1/9. Let's check if x and z found here satisfy this. x = 5/36, z = -1/36. So 5/36 -1/36 = 4/36 = 1/9. Yes, that works. So that's correct. So z = -1/36. So 1/c = -1/36 => c = -36. Hmm. So c is negative. Interesting. Let's see if that's okay.So then, moving on. We need to find c/(ab). Let's compute c, a, b.We have x = 1/a = 5/36 => a = 36/5.y = 1/b =7/36 => b = 36/7.z = 1/c = -1/36 => c = -36.So ab = (36/5)*(36/7) = (36*36)/(5*7) = 1296/35.Then c/(ab) = (-36)/(1296/35) = (-36)*(35/1296). Let's compute that.First, simplify 1296/36. 1296 ÷ 36 = 36. So 36 ÷ 1296 is 1/36. But here, it's (-36) * 35 / 1296.So 36 cancels with 1296. 1296 ÷ 36 = 36. So:(-36) * 35 / 1296 = (-35)/36.So c/(ab) = -35/36.Wait, but the question is phrased as "then c/(ab) =", and it's possible they just want the numerical value, regardless of sign. But I need to check if my answer is correct. Let's verify all steps again.First, x = 5/36, so a = 36/5.y =7/36, so b = 36/7.z = -1/36, so c = -36.Check first equation: ab/(a + b).a = 36/5, b = 36/7.ab = (36/5)(36/7) = 1296/35.a + b = 36/5 + 36/7 = 36(7 +5)/35 = 36*12/35 = 432/35.Then ab/(a + b) = (1296/35) / (432/35) = 1296/432 = 3. Correct. The first equation is satisfied.Second equation: bc/(b + c).b = 36/7, c = -36.bc = (36/7)(-36) = -1296/7.b + c = 36/7 - 36 = 36/7 - 252/7 = (-216)/7.Then bc/(b + c) = (-1296/7)/(-216/7) = 1296/216 = 6. Correct. The second equation is satisfied.Third equation: ac/(a + c).a = 36/5, c = -36.ac = (36/5)(-36) = -1296/5.a + c = 36/5 - 36 = 36/5 - 180/5 = (-144)/5.Then ac/(a + c) = (-1296/5)/(-144/5) = 1296/144 = 9. Correct. Third equation is satisfied.So all equations check out. Therefore, the values of a, b, c are correct. Then c/(ab) is (-36)/(1296/35) which is (-36)*(35/1296) = (-35)/36. So the answer is -35/36. But let me check the problem statement again. It just says "then c/(ab) =". It doesn't specify any constraints on the variables. So, if they allow negative numbers, then this is the answer.But maybe I made a mistake in the substitution? Let's check z again. z = -1/36, so c = -36. Then, in the third equation, ac/(a + c) = 9. Plugging in a = 36/5 and c = -36:ac = (36/5)(-36) = -1296/5.a + c = 36/5 - 36 = 36/5 - 180/5 = -144/5.Then ac/(a + c) = (-1296/5)/(-144/5) = 1296/144 = 9. Correct.So all equations are satisfied. Therefore, even though c is negative, the answer is -35/36. So unless there's a mistake in the calculation steps, which I don't see, that's the answer. Let me check once again the computation of c/(ab):c = -36ab = (36/5)(36/7) = (36*36)/(5*7) = 1296/35Then c/(ab) = (-36) / (1296/35) = (-36) * (35/1296) = (-36*35)/1296Simplify numerator and denominator by 36: numerator becomes -35, denominator 36. So yes, -35/36.So that's the answer. Therefore, the final answer is boxed{-dfrac{35}{36}}.But wait, the problem is given in a math competition or something, maybe the answer is expected to be positive? Let me check again the equations. If c is negative, then ac and bc are negative, but the ratios (ac)/(a + c) and (bc)/(b + c) are positive. So denominators a + c and b + c must also be negative. Let's check that.For example, a + c = 36/5 + (-36) = 36/5 - 180/5 = -144/5, which is negative. Similarly, b + c = 36/7 - 36 = 36/7 - 252/7 = -216/7, which is negative. So negative over negative gives positive. So that's okay. So c being negative is acceptable here. So the answer is indeed negative. So, unless there was a typo in the problem, this is correct.But let me check the problem statement again. The user wrote:"If (ab)/(a+b)=3, (bc)/(b+c)=6, (ac)/(a+c)=9, then c/(ab)= "So, given that, the answer is -35/36. So unless there is a miscalculation here, which I don't see, that's the answer. Therefore, the answer is -35/36.Final Answerboxed{-dfrac{35}{36}}</think>Given the equations:1. (frac{ab}{a + b} = 3)2. (frac{bc}{b + c} = 6)3. (frac{ac}{a + c} = 9)We need to find (frac{c}{ab}).First, we introduce the substitutions (x = frac{1}{a}), (y = frac{1}{b}), and (z = frac{1}{c}). This transforms the given equations into a system of linear equations:1. (x + y = frac{1}{3})2. (y + z = frac{1}{6})3. (x + z = frac{1}{9})We solve this system step-by-step:1. Subtract the third equation from the first: [ (x + y) - (x + z) = frac{1}{3} - frac{1}{9} ] Simplifying, we get: [ y - z = frac{2}{9} ]2. Using the second equation (y + z = frac{1}{6}), we add and subtract the equations: [ (y - z) + (y + z) = frac{2}{9} + frac{1}{6} ] Simplifying, we find: [ 2y = frac{7}{18} implies y = frac{7}{36} ]3. Substitute (y = frac{7}{36}) back into the first equation: [ x + frac{7}{36} = frac{1}{3} implies x = frac{1}{3} - frac{7}{36} = frac{12}{36} - frac{7}{36} = frac{5}{36} ]4. Substitute (y = frac{7}{36}) into the second equation: [ frac{7}{36} + z = frac{1}{6} implies z = frac{1}{6} - frac{7}{36} = frac{6}{36} - frac{7}{36} = -frac{1}{36} ]Thus, we have (x = frac{5}{36}), (y = frac{7}{36}), and (z = -frac{1}{36}). Converting back to (a), (b), and (c):- (a = frac{1}{x} = frac{36}{5})- (b = frac{1}{y} = frac{36}{7})- (c = frac{1}{z} = -36)Finally, we calculate (frac{c}{ab}):[ab = left(frac{36}{5}right) left(frac{36}{7}right) = frac{1296}{35}][frac{c}{ab} = frac{-36}{frac{1296}{35}} = -36 times frac{35}{1296} = -frac{35}{36}]Thus, the final answer is (boxed{-dfrac{35}{36}}).]

🔑:Answer: -frac{35}{36}

🧠:[<think>Okay, so the problem is asking for the range of the real number ( a ) such that the proposition "There exists an ( x in mathbb{R} ) such that ( x^2 + (a-1)x + 1 leq 0 )" is false. Hmm, that means we need to find all values of ( a ) for which the quadratic equation ( x^2 + (a-1)x + 1 leq 0 ) has no real solutions. First, I remember that a quadratic equation ( ax^2 + bx + c = 0 ) has no real solutions if its discriminant is negative. But here, the inequality is ( leq 0 ). So, if the quadratic never touches or goes below the x-axis, then the inequality would be false for all real ( x ). Therefore, the quadratic should be always positive. Wait, so if the quadratic ( x^2 + (a-1)x + 1 ) is always positive for all real ( x ), then the original statement is false. Therefore, the condition we need is that the quadratic is always positive. For a quadratic to be always positive, the leading coefficient must be positive, and the discriminant must be negative. Let me check: the leading coefficient here is 1 (coefficient of ( x^2 )), which is positive, so that's already satisfied. Then, the discriminant must be negative. The discriminant ( D ) of ( x^2 + (a-1)x + 1 ) is ( D = (a - 1)^2 - 4 times 1 times 1 ). So, ( D = (a - 1)^2 - 4 ). For the quadratic to be always positive, we need ( D < 0 ). Therefore:( (a - 1)^2 - 4 < 0 )Let me solve this inequality:( (a - 1)^2 < 4 )Taking square roots on both sides, we get:( |a - 1| < 2 )Which translates to:( -2 < a - 1 < 2 )Adding 1 to all parts:( -2 + 1 < a < 2 + 1 )Simplifies to:( -1 < a < 3 )Therefore, the range of ( a ) should be ( (-1, 3) ). Let me verify this.Suppose ( a = 0 ), which is in the range. Then the quadratic becomes ( x^2 - x + 1 ). The discriminant is ( (-1)^2 - 4 times 1 times 1 = 1 - 4 = -3 < 0 ), so the quadratic is always positive. Correct.If ( a = 4 ), which is outside the range. Then the quadratic is ( x^2 + 3x + 1 ). The discriminant is ( 9 - 4 = 5 > 0 ), so there are real roots. Hence, the quadratic will take negative values between the roots. So the original statement would be true. Therefore, ( a = 4 ) is excluded, which is correct.Similarly, if ( a = -2 ), outside the range, the quadratic is ( x^2 - 3x + 1 ). Discriminant is ( 9 - 4 = 5 > 0 ), so again real roots and the quadratic takes negative values. So original statement is true, which is why ( a = -2 ) is excluded. Therefore, the range of ( a ) is indeed ( (-1, 3) ). But wait, let me double-check the discriminant calculation. The quadratic is ( x^2 + (a - 1)x + 1 ). So the coefficients are:( a_{text{quad}} = 1 ), ( b_{text{quad}} = (a - 1) ), ( c_{text{quad}} = 1 ).Therefore, discriminant ( D = b^2 - 4ac = (a - 1)^2 - 4(1)(1) = (a - 1)^2 - 4 ). Correct.So solving ( (a - 1)^2 - 4 < 0 ):( (a - 1)^2 < 4 )Which is equivalent to ( |a - 1| < 2 ), so ( -2 < a - 1 < 2 ), leading to ( -1 < a < 3 ). That's correct.Hence, the range is ( a in (-1, 3) ). Alternatively, if the problem had asked for when the proposition is true, we would have considered the quadratic having at least one real solution where the quadratic is less than or equal to zero, which would occur when the discriminant is non-negative. But since the proposition is false, we need the opposite: the quadratic is always positive, hence discriminant negative. Therefore, the answer should be ( -1 < a < 3 ).Final AnswerThe range of the real number ( a ) is boxed{(-1, 3)}.</think>To determine the range of the real number ( a ) such that the proposition "There exists an ( x in mathbb{R} ) such that ( x^2 + (a-1)x + 1 leq 0 )" is false, we need to ensure that the quadratic expression ( x^2 + (a-1)x + 1 ) is always positive for all ( x in mathbb{R} ).For a quadratic expression ( ax^2 + bx + c ) to be always positive, the leading coefficient must be positive and the discriminant must be negative. Here, the leading coefficient is 1, which is positive. Therefore, we need the discriminant to be negative.The discriminant ( D ) of the quadratic expression ( x^2 + (a-1)x + 1 ) is given by:[D = (a - 1)^2 - 4 times 1 times 1 = (a - 1)^2 - 4]For the quadratic to be always positive, we require:[(a - 1)^2 - 4 < 0]Solving this inequality:[(a - 1)^2 < 4]Taking the square root of both sides:[|a - 1| < 2]This translates to:[-2 < a - 1 < 2]Adding 1 to all parts:[-1 < a < 3]Thus, the range of the real number ( a ) is (boxed{(-1, 3)}).,<think>Okay, so I need to figure out the range of the real number ( a ) for which the proposition "There exists an ( x in mathbb{R} ) such that ( x^2 + (a-1)x + 1 leq 0 )" is false. That means the quadratic equation ( x^2 + (a-1)x + 1 leq 0 ) should have no real solutions. Hmm, right. So if the quadratic never goes below or equal to zero, that means it's always positive. So the quadratic ( x^2 + (a-1)x + 1 ) is always greater than zero for all real ( x ). First, I remember that a quadratic ( ax^2 + bx + c ) is always positive if the leading coefficient is positive and the discriminant is negative. In this case, the quadratic is ( x^2 + (a-1)x + 1 ), so the leading coefficient is 1, which is positive. So that condition is satisfied. So then, the discriminant must be negative for there to be no real roots. The discriminant ( D ) of a quadratic ( ax^2 + bx + c ) is ( D = b^2 - 4ac ). Here, ( a ) is 1, ( b ) is ( (a-1) ), and ( c ) is 1. Wait, but in the standard quadratic formula, the coefficients are ( ax^2 + bx + c ), so here, the coefficient of ( x ) is ( (a-1) ), so ( b = (a - 1) ). Therefore, the discriminant ( D ) would be ( (a - 1)^2 - 4 * 1 * 1 ). So ( D = (a - 1)^2 - 4 ). For the quadratic to have no real roots, we need ( D < 0 ). Therefore, ( (a - 1)^2 - 4 < 0 ). Solving this inequality should give the range of ( a ).Let me solve ( (a - 1)^2 < 4 ). Taking square roots on both sides, we get ( |a - 1| < 2 ). This implies that ( -2 < a - 1 < 2 ). Adding 1 to all parts, we get ( -2 + 1 < a < 2 + 1 ), so ( -1 < a < 3 ). Wait, but let me double-check. If the discriminant is less than zero, then there are no real roots, which would mean the quadratic is always positive since the leading coefficient is positive. Therefore, the original proposition that there exists an ( x ) such that the quadratic is less than or equal to zero would be false. So that would mean the range of ( a ) is ( (-1, 3) ). But let me make sure I didn't make a mistake in solving the inequality. Let's start again:( (a - 1)^2 - 4 < 0 )( (a - 1)^2 < 4 )Taking square roots, since square root is a monotonic function for non-negative numbers, so the inequality direction remains the same when taking square roots. So:( |a - 1| < 2 )Which translates to:( -2 < a - 1 < 2 )Adding 1:( -2 + 1 < a < 2 + 1 )( -1 < a < 3 )Yes, that seems correct. So the range of ( a ) is ( (-1, 3) ). Wait, but just to make sure, let's test a value in this interval and outside to check. Let's pick ( a = 0 ), which is inside the interval. Then the quadratic becomes ( x^2 + (-1)x + 1 ). Let's compute its discriminant: ( (-1)^2 - 4*1*1 = 1 - 4 = -3 < 0 ). So no real roots, and since the leading coefficient is positive, it's always positive. So the proposition is false here. Now, take ( a = 4 ), which is outside the interval. The quadratic is ( x^2 + (4 - 1)x + 1 = x^2 + 3x + 1 ). Discriminant is ( 9 - 4 = 5 > 0 ). So it has two real roots, meaning the quadratic crosses the x-axis, so there are x-values where it is negative. Therefore, the proposition is true here. Similarly, take ( a = -2 ), outside the interval. Quadratic becomes ( x^2 + (-2 -1)x + 1 = x^2 -3x + 1 ). Discriminant is ( 9 - 4 = 5 > 0 ). Again, two real roots, so the quadratic takes negative values. Therefore, the proposition is true here as well. Another test with ( a = 2 ), which is inside the interval? Wait, 2 is inside (-1,3). Quadratic is ( x^2 + (2 -1)x + 1 = x^2 + x + 1 ). Discriminant is ( 1 - 4 = -3 < 0 ). So no real roots, quadratic is always positive. Proposition is false. Correct. And ( a = -0.5 ), which is inside the interval. Quadratic is ( x^2 + (-0.5 -1)x +1 = x^2 - 1.5x +1 ). Discriminant: ( (1.5)^2 -4*1*1 = 2.25 -4 = -1.75 < 0 ). So no real roots. Correct. Therefore, the range is indeed ( -1 < a < 3 ), so the answer should be ( (-1, 3) ). But let me check again if the problem is stated correctly. The proposition is "There exists an ( x in mathbb{R} ) such that ( x^2 + (a -1)x +1 leq 0 )" is false. So, if the quadratic never is less than or equal to zero, which requires that the quadratic is always positive. So, discriminant negative. Therefore, as we did. So, yeah, the answer is ( a ) in ( (-1, 3) ). But maybe the problem wants the answer in interval notation? Then writing ( (-1, 3) ). Alternatively, using inequalities: ( -1 < a < 3 ). Both are correct. Since the question says "the range of the real number ( a )", so either interval notation or inequalities are okay, but the user probably wants the boxed answer. Wait, the user instruction says: "put your final answer within boxed{}". So, probably boxed interval notation. So, boxed{(-1, 3)}. But let me just confirm once again. Suppose ( a = -1 ). Then the quadratic is ( x^2 + (-1 -1)x +1 = x^2 -2x +1 = (x -1)^2 ). So, this is a perfect square, which is always non-negative, but it touches zero at x=1. Therefore, the proposition "There exists an x such that ( x^2 + (a-1)x +1 leq 0 )" would be true, because at x=1, it's equal to zero. Therefore, when ( a = -1 ), the proposition is true, so ( a = -1 ) is excluded from the range. Similarly, ( a = 3 ): quadratic is ( x^2 + (3 -1)x +1 = x^2 + 2x +1 = (x +1)^2 ). Again, it's a perfect square, zero at x = -1, so the proposition is true. Therefore, ( a ) must be strictly greater than -1 and strictly less than 3. So the interval is open, which matches our previous result. Therefore, the answer is boxed{(-1, 3)}.Final AnswerThe range of the real number ( a ) is boxed{(-1, 3)}.</think>To determine the range of the real number ( a ) for which the proposition "There exists an ( x in mathbb{R} ) such that ( x^2 + (a-1)x + 1 leq 0 )" is false, we need to ensure that the quadratic expression ( x^2 + (a-1)x + 1 ) is always positive for all real ( x ).A quadratic polynomial ( ax^2 + bx + c ) is always positive if its leading coefficient is positive and its discriminant is negative. Here, the quadratic is ( x^2 + (a-1)x + 1 ), so the leading coefficient is 1 (which is positive). Therefore, we need the discriminant to be negative.The discriminant ( D ) of the quadratic equation ( x^2 + (a-1)x + 1 ) is given by:[D = (a - 1)^2 - 4 cdot 1 cdot 1][D = (a - 1)^2 - 4]For the quadratic to have no real roots, the discriminant must be negative:[(a - 1)^2 - 4 < 0]Solving this inequality:[(a - 1)^2 < 4]Taking the square root of both sides:[|a - 1| < 2]This implies:[-2 < a - 1 < 2]Adding 1 to all parts of the inequality:[-1 < a < 3]Thus, the range of the real number ( a ) is (boxed{(-1, 3)}).]

🔑:Since the proposition states "There exists an x in mathbb{R} such that x^2 + (a-1)x + 1 leq 0" and we know this proposition is false, it means there are no real values of x that can satisfy the inequality. For such a quadratic inequality to be always positive, the corresponding quadratic equation x^2 + (a-1)x + 1 = 0 must not have any real roots. To ensure this, the discriminant of the equation must be less than zero.The discriminant Delta of the quadratic equation ax^2 + bx + c = 0 is given by Delta = b^2 - 4ac. Applying this to our equation, we have:Delta = (a-1)^2 - 4 cdot 1 cdot 1 = a^2 - 2a + 1 - 4.For the quadratic equation to have no real roots, the discriminant must be negative:Delta < 0a^2 - 2a + 1 - 4 < 0a^2 - 2a - 3 < 0.This is a quadratic inequality which we can factor as:(a-3)(a+1) < 0.By examining the critical points and the behavior of the quadratic, we find that the inequality holds true for:-1 < a < 3.Therefore, the range of the real number a for which the original proposition is false is (-1, 3).[boxed{-1 < a < 3}]

🧠:[<think>Okay, let's try to solve this geometry problem. So, in triangle ABC, we have an altitude AH and an angle bisector AL. The segments BH, HL, and LC are in the ratio 1:1:4, and side AB is equal to 1. We need to find the lengths of the other sides, AC and BC. Hmm, let's see.First, let me visualize the triangle. Let me sketch it mentally. Triangle ABC with vertex A at the top, BC as the base. AH is the altitude from A to BC, so H is the foot of the altitude. AL is the angle bisector from A to BC. The points B, H, L, and C are on BC in that order, right? Because BH:HL:LC = 1:1:4. So BH is the first segment, then HL, then LC. So H is between B and L, and L is between H and C. Wait, but if the ratio is BH:HL:LC=1:1:4, then starting from B, we go 1 part to H, another 1 part to L, and then 4 parts to C. So the entire BC length would be 1+1+4=6 parts. So BH=1k, HL=1k, LC=4k, so BC=6k for some k.But wait, AL is the angle bisector. So the angle bisector theorem comes into play here. The angle bisector theorem states that the angle bisector divides the opposite side into segments proportional to the adjacent sides. That is, BL / LC = AB / AC. Wait, but here L is the point where the angle bisector meets BC. But in the problem, we have AL as the bisector. So yes, L is on BC such that AL bisects angle A. Therefore, BL / LC = AB / AC. Given that AB is 1, so BL / LC = 1 / AC. Therefore, AC = LC / BL.Wait, but we have BH:HL:LC = 1:1:4. So from B to H is 1, H to L is 1, L to C is 4. Therefore, BL is BH + HL = 1 + 1 = 2 parts, and LC is 4 parts. So BL / LC = 2 / 4 = 1 / 2. Therefore, by the angle bisector theorem, BL / LC = AB / AC => 1/2 = 1 / AC => AC = 2. So AC is 2. Wait, that seems straightforward. So AC would be 2. Then BC is 6 parts, so BH=1k, HL=1k, LC=4k. So BC=6k, but we need to find BC. But how do we find k?Wait, but we also have the altitude AH. Since AH is the altitude, we can use the Pythagorean theorem in triangles ABH and AHC. Let me denote AH as h, BH as k, HL as k, and LC as 4k. So the total BC is 6k. Then, in triangle ABH, which is right-angled at H, AB^2 = AH^2 + BH^2. So AB is given as 1, so 1 = h^2 + k^2. Similarly, in triangle AHC, which is right-angled at H, AC^2 = AH^2 + HC^2. But HC is HL + LC = k + 4k = 5k. Wait, no. Wait, HC is from H to C. Since H is between B and L, and L is between H and C, then HC would be HL + LC = 1k + 4k = 5k. So HC = 5k. Therefore, AC^2 = h^2 + (5k)^2. But earlier, we found AC = 2, right? So 2^2 = h^2 + (5k)^2 => 4 = h^2 + 25k^2. But we also have from triangle ABH that h^2 = 1 - k^2. So substituting into the equation for AC:4 = (1 - k^2) + 25k^2 => 4 = 1 + 24k^2 => 24k^2 = 3 => k^2 = 3 / 24 = 1 / 8 => k = 1 / (2√2) = √2 / 4. Hmm, okay. So k is √2 / 4. Then BH = k = √2 / 4, HL = k = √2 / 4, LC = 4k = √2. Then BC = 6k = 6*(√2 / 4) = (3√2)/2. So BC is (3√2)/2. So then AC is 2, BC is (3√2)/2. So the sides are AB=1, AC=2, BC=3√2 / 2. Let me check if that makes sense.Wait, but let's verify using another method. Let's see. If AL is the angle bisector, then by the angle bisector theorem, as we did before, BL / LC = AB / AC. Since BL is 2k and LC is 4k, so 2k / 4k = 1 / 2, so 1 / 2 = AB / AC. AB is 1, so AC is 2. So that part checks out. Then using the altitude, we found k = √2 / 4, so BC = 6k = (6√2)/4 = (3√2)/2. So BC is (3√2)/2. So the other sides are AC=2 and BC=(3√2)/2. That seems correct.Wait, but let's check if the altitude AH is consistent. If BH is √2 / 4, then AH^2 + BH^2 = AB^2. So AH^2 = 1 - ( (√2 / 4)^2 ) = 1 - (2 / 16) = 1 - (1/8) = 7/8. Therefore, AH = √(7/8) = (√14)/4. Then, HC is HL + LC = √2 / 4 + 4*(√2 / 4) = √2 / 4 + √2 = (5√2)/4. Then AC^2 should be AH^2 + HC^2. Let's compute that: (7/8) + (25*2)/16 = 7/8 + 50/16 = 7/8 + 25/8 = 32/8 = 4. Which is correct because AC is 2. So that works out.So, seems like the calculations are correct. Therefore, the lengths of the other sides are AC=2 and BC= (3√2)/2.But wait, just to make sure there's no mistake here, let's go through the steps again.1. Given triangle ABC with altitude AH and angle bisector AL. Ratios BH:HL:LC=1:1:4. AB=1. Find AC and BC.2. Let BH = x, HL = x, LC =4x. Therefore, BC = 6x.3. By angle bisector theorem, BL / LC = AB / AC. BL = BH + HL = x + x = 2x. LC =4x. Therefore, 2x / 4x = 1 / AC => 1/2 = 1 / AC => AC=2. So AC=2.4. Now, using altitude AH. In right triangles ABH and AHC: - AB^2 = AH^2 + BH^2 => 1 = AH^2 + x^2 => AH^2 = 1 - x^2. - AC^2 = AH^2 + HC^2 => 4 = AH^2 + (5x)^2.5. Substitute AH^2 from first equation into second: 4 = (1 - x^2) + 25x^2 => 4 = 1 + 24x^2 => 24x^2 = 3 => x^2 = 1/8 => x = √(1/8) = (√2)/4.6. Therefore, BC =6x =6*(√2)/4 = (3√2)/2.All steps check out. Therefore, AC=2 and BC= (3√2)/2.Alternatively, another way to approach it could be using coordinate geometry. Let me try that as a verification.Let me place point B at (0,0). Then, since AH is the altitude from A to BC, and H is on BC. Let me let BC lie along the x-axis. So point B is at (0,0), H is at (x,0), L is at (2x,0), and C is at (6x,0). Wait, wait, the ratio is BH:HL:LC=1:1:4. So BH=1k, HL=1k, LC=4k, so BC=6k. So starting from B at 0, H is at k, L is at 2k, C is at 6k. Then AH is the altitude, so A is at (k, h) because the foot of the altitude is at H which is at (k,0). Wait, if we place B at (0,0), then BC is along the x-axis, H is at (k,0), L at (2k,0), C at (6k,0). Then point A is at (k, h) since AH is the altitude. Then AB is the distance from A(k,h) to B(0,0): sqrt((k-0)^2 + (h-0)^2) = sqrt(k^2 + h^2) =1. Similarly, AC is the distance from A(k,h) to C(6k,0): sqrt((6k -k)^2 + (0 - h)^2)=sqrt(25k^2 + h^2)=2.So we have two equations:1. sqrt(k^2 + h^2) =1 => k^2 + h^2 =1.2. sqrt(25k^2 + h^2)=2 =>25k^2 + h^2=4.Subtracting first equation from second: 24k^2=3 => k^2=1/8 => k=1/(2√2)=√2/4. Then h^2=1 -1/8=7/8 => h=√14/4.So point A is at (√2/4, √14/4). Then, AL is the angle bisector from A to BC. Let's check if point L is indeed at (2k,0). So point L is at (2k,0)= (2*(√2/4),0)= (√2/2,0). Now, angle bisector AL goes from A(√2/4, √14/4) to L(√2/2,0). Let's check if this is indeed the angle bisector. Alternatively, since we already used the angle bisector theorem to find AC=2, which is consistent with the coordinates.Alternatively, the coordinates method gives the same results. Therefore, BC is 6k=6*(√2/4)= (3√2)/2. Therefore, the sides are AB=1, AC=2, BC=3√2/2. So that confirms the answer.Another way to think about this problem is using the properties of angle bisectors and the Pythagorean theorem. We found AC via the angle bisector theorem, then used the altitude to form two right triangles, which allowed us to solve for the variable k, leading to BC. Everything seems consistent.Therefore, the lengths of the other sides are AC=2 and BC= (3√2)/2.Final AnswerThe lengths of the other sides are boxed{2} and boxed{dfrac{3sqrt{2}}{2}}.</think>Given triangle (ABC) with altitude (AH) and angle bisector (AL). The segments (BH), (HL), and (LC) are in the ratio (1:1:4), and side (AB) is equal to 1. We need to find the lengths of the other sides (AC) and (BC).1. Identify the segments: - Let (BH = x), (HL = x), and (LC = 4x). - Therefore, (BC = BH + HL + LC = x + x + 4x = 6x).2. Apply the Angle Bisector Theorem: - The angle bisector (AL) divides (BC) such that (frac{BL}{LC} = frac{AB}{AC}). - Here, (BL = BH + HL = x + x = 2x) and (LC = 4x). - Thus, (frac{BL}{LC} = frac{2x}{4x} = frac{1}{2} = frac{AB}{AC}). - Given (AB = 1), we get (frac{1}{AC} = frac{1}{2}), so (AC = 2).3. Use the Pythagorean Theorem: - In right triangle (ABH), (AB^2 = AH^2 + BH^2): [ 1 = AH^2 + x^2 implies AH^2 = 1 - x^2 ] - In right triangle (AHC), (AC^2 = AH^2 + HC^2): [ 4 = AH^2 + (5x)^2 ] - Substitute (AH^2) from the first equation into the second: [ 4 = (1 - x^2) + 25x^2 implies 4 = 1 + 24x^2 implies 24x^2 = 3 implies x^2 = frac{1}{8} implies x = frac{sqrt{2}}{4} ]4. Calculate (BC): - (BC = 6x = 6 times frac{sqrt{2}}{4} = frac{3sqrt{2}}{2}).Thus, the lengths of the other sides are (boxed{2}) and (boxed{dfrac{3sqrt{2}}{2}}).,<think>Okay, so I have this geometry problem here. Let me try to figure it out step by step. The problem is about triangle ABC where an altitude AH and a bisector AL are drawn. The ratios of BH:HL:LC are given as 1:1:4, and the length of side AB is 1. I need to find the lengths of the other sides of triangle ABC. Alright, let's start by visualizing the triangle. First, let me sketch triangle ABC. Let me label the vertices A, B, and C. Since AH is the altitude from A to BC, that means AH is perpendicular to BC. Then, AL is the bisector of angle A, so it splits angle BAC into two equal angles. The points H and L are both on side BC, and the segments BH, HL, and LC are in the ratio 1:1:4. So, BH is one part, HL is another part, and LC is four parts. Since BH:HL:LC = 1:1:4, the total parts are 1 + 1 + 4 = 6 parts. So, the entire length of BC is divided into six parts, with BH being 1, HL being 1, and LC being 4. Given that AB is 1, and we need to find the lengths of AC and BC. Hmm. Let me denote some variables. Let's let the length BH = x, HL = x, and LC = 4x. So, BC = BH + HL + LC = x + x + 4x = 6x. So, BC = 6x. So, if I can find x, then I can find BC. Also, we need to find AC. Since AL is the angle bisector, by the Angle Bisector Theorem, the ratio of the adjacent sides is equal to the ratio of the segments created by the bisector on the opposite side. The Angle Bisector Theorem states that if a bisector of an angle of a triangle divides the opposite side into segments proportional to the adjacent sides. So, in this case, since AL is the bisector of angle A, it should divide BC into segments proportional to AB and AC. So, BL / LC = AB / AC. Wait, but here, L is a point on BC such that BL = BH + HL = x + x = 2x, and LC = 4x. So, BL / LC = (2x)/(4x) = 1/2. Therefore, according to the Angle Bisector Theorem, AB / AC = BL / LC = 1/2. Since AB is given as 1, then 1 / AC = 1/2, so AC = 2. Wait, is that correct? So, AC would be 2? Let me check again. Yes, Angle Bisector Theorem says BL / LC = AB / AC. So, if BL is 2x and LC is 4x, then BL/LC = 2x/4x = 1/2. So, AB/AC = 1/2. Since AB is 1, then AC must be 2. So, AC = 2. So, that gives me AC. Then, BC is 6x. So, if I can find x, then I can find BC. But how?Since AH is the altitude from A to BC, then AH is perpendicular to BC. Therefore, triangle ABH and triangle ACH are right-angled triangles. Let's consider triangle ABH. In triangle ABH, AH is the altitude, so by the Pythagorean theorem, AB² = AH² + BH². Similarly, in triangle AHC, AC² = AH² + HC². But HC is HL + LC = x + 4x = 5x. Wait, no. Wait, H is the foot of the altitude from A to BC, so H is a point on BC such that AH is perpendicular to BC. However, the problem states that BH:HL:LC = 1:1:4. So, H is between B and L, and L is between H and C. So, the order on BC is B, H, L, C. Therefore, BH = x, HL = x, LC = 4x. Therefore, the total BC is 6x. So, H is x distance from B, then L is another x distance from H, and then C is 4x from L. So, HC would be HL + LC? Wait, no. Wait, H is a point on BC; if we have BH = x, then the rest of BC from H is HL + LC = x + 4x = 5x. So, HC = HL + LC = 5x. But wait, HC is the length from H to C, which is HL + LC = x + 4x = 5x. Similarly, BH is x. So, in triangle ABH, right-angled at H, we have AB² = AH² + BH². So, AB is given as 1, BH is x. So, 1² = AH² + x². So, AH² = 1 - x². Similarly, in triangle AHC, which is right-angled at H, we have AC² = AH² + HC². AC we found to be 2, so 2² = AH² + (5x)^2. So, 4 = AH² + 25x². But we know AH² is 1 - x² from the previous equation. So, substituting, 4 = (1 - x²) + 25x². That simplifies to 4 = 1 + 24x². Therefore, 24x² = 3, so x² = 3/24 = 1/8. Therefore, x = 1/(2√2) or x = √2/4. Since x is a length, we can take the positive value. So, x = √2/4. Then, BC = 6x = 6*(√2/4) = (3√2)/2. Wait, so BC would be (3√2)/2. Let me check if this makes sense. Let me verify the calculations again. We have AB = 1, which is given. Then, AC = 2 from the Angle Bisector Theorem. Then, BH = x = √2/4. Then, AH² = 1 - x² = 1 - ( (√2/4)^2 ) = 1 - (2/16) = 1 - 1/8 = 7/8. So, AH = √(7/8) = (√14)/4. Then, HC = 5x = 5*(√2/4) = (5√2)/4. Then, in triangle AHC, AC² should equal AH² + HC². Let's check that: AH² + HC² = 7/8 + (25*2)/16 = 7/8 + 50/16 = 7/8 + 25/8 = 32/8 = 4. Which is equal to AC² (since AC = 2, 2² = 4). So, that checks out. Similarly, in triangle ABH, AB² = 1, and AH² + BH² = 7/8 + 2/16 = 7/8 + 1/8 = 8/8 = 1. Which also checks out. So, the calculations seem correct. Therefore, BC is 6x = 6*(√2/4) = (3√2)/2, and AC is 2. So, the sides are AB = 1, AC = 2, and BC = (3√2)/2. Wait, but let me check if there's another approach or if I missed something. Because sometimes with ratios and angle bisectors, there might be other relationships. Let me verify again the application of the Angle Bisector Theorem. The Angle Bisector Theorem states that the angle bisector divides the opposite side into segments proportional to the adjacent sides. So, in triangle ABC, angle bisector AL divides BC into BL and LC such that BL/LC = AB/AC. In this problem, the points on BC are in the order B, H, L, C, with BH:HL:LC = 1:1:4. Therefore, BL = BH + HL = 1 + 1 = 2 parts, and LC = 4 parts. So, BL/LC = 2/4 = 1/2. Therefore, AB/AC = 1/2. Since AB is 1, then AC must be 2. That seems correct. Then, using the altitude AH, we found x by relating the lengths in the right triangles ABH and AHC. The key was expressing AH² in terms of x from triangle ABH and then substituting into the equation from triangle AHC. That led us to solve for x, which then gave BC. So, unless there's an error in my ratio understanding or in the application of the theorems, this should be correct. Let me just check the ratio once more. Given BH:HL:LC = 1:1:4. So, BH = x, HL = x, LC = 4x. Then, BL = BH + HL = x + x = 2x, and LC = 4x. So, BL/LC = 2x/4x = 1/2. So, AB/AC = 1/2. AB is 1, so AC = 2. That's correct. Then, BC is BH + HL + LC = x + x + 4x = 6x. Then, using the altitude, we found x = √2/4. So, BC = 6*(√2)/4 = (3√2)/2. Therefore, the other sides are AC = 2 and BC = (3√2)/2. That seems to be the answer. Let me just check if there's a different interpretation of the problem. Wait, could the order of the points on BC be different? The problem says BH:HL:LC = 1:1:4. So, the segments are BH, HL, LC. So, starting from B, first BH, then HL, then LC. So, the order is B, H, L, C. So, that's correct. So, H is between B and L, and L is between H and C. Therefore, the foot of the altitude H is between B and L, and the angle bisector L is between H and C. But in that case, how come the angle bisector L is further away from C? Wait, the angle bisector usually divides the opposite side in the ratio of the adjacent sides. Since AB is 1 and AC is 2, the angle bisector should divide BC into a ratio of AB to AC, which is 1:2. But in our problem, the ratio BL/LC is 1/2. Wait, BL is 2x, LC is 4x, so BL/LC is 1/2, which is the same as AB/AC (1/2). So, that's consistent. Therefore, even though L is closer to C, because the ratio is 1:2, which is consistent with the Angle Bisector Theorem. Therefore, all the steps check out. So, the lengths are AC = 2 and BC = 3√2/2. So, the final answer should be AC = 2 and BC = (3√2)/2. Wait, but let me check once more with coordinates. Maybe that can help solidify the answer. Let's place point B at the origin (0,0), and point H somewhere along the x-axis. Let me assign coordinates. Let me set BC along the x-axis for simplicity. Wait, but AH is the altitude, so AH is perpendicular to BC. If I place BC on the x-axis, then AH would be vertical. Let me try that. Let me set point B at (0,0). Then, since BH = x, point H is at (x, 0). Then, HL = x, so point L is at (x + x, 0) = (2x, 0). Then, LC = 4x, so point C is at (2x + 4x, 0) = (6x, 0). Therefore, BC is from (0,0) to (6x,0). Then, the altitude AH is drawn from A to H, which is at (x,0). Since AH is perpendicular to BC, which is along the x-axis, AH is vertical. Therefore, point A has coordinates (x, h), where h is the height AH. Then, AB is the distance from (x, h) to (0,0), which is sqrt((x - 0)^2 + (h - 0)^2) = sqrt(x² + h²). Given that AB = 1, so sqrt(x² + h²) = 1. Therefore, x² + h² = 1. Similarly, AC is the distance from (x, h) to (6x, 0), which is sqrt((6x - x)^2 + (0 - h)^2) = sqrt((5x)^2 + h²) = sqrt(25x² + h²). We found that AC = 2, so sqrt(25x² + h²) = 2. Therefore, 25x² + h² = 4. From AB, we have h² = 1 - x². Substitute into the equation for AC: 25x² + (1 - x²) = 4. Simplify: 24x² + 1 = 4. Therefore, 24x² = 3, so x² = 3/24 = 1/8. So, x = 1/(2√2) = √2/4. Then, h² = 1 - x² = 1 - 1/8 = 7/8, so h = sqrt(7/8) = (√14)/4. Therefore, the coordinates of point A are (x, h) = (√2/4, √14/4). Then, coordinates of points: B (0,0), H (√2/4, 0), L (2*(√2/4), 0) = (√2/2, 0), and C (6*(√2/4), 0) = (3√2/2, 0). Now, let's check the angle bisector AL. AL is the bisector of angle A, so it should split angle BAC into two equal angles. Let's verify if point L is indeed the point where the angle bisector meets BC. The coordinates of point L are (√2/2, 0). To check if AL is the angle bisector, we can use the Angle Bisector Theorem, which we already did, but let's verify using coordinates. The vector approach or slope approach might be complicated, but since we already verified the ratio, maybe it's sufficient. Alternatively, we can compute the angles. Alternatively, compute the direction vectors of AL and see if it bisects the angle between AB and AC. Coordinates of A: (√2/4, √14/4). Coordinates of L: (√2/2, 0). So, vector AL is (√2/2 - √2/4, 0 - √14/4) = (√2/4, -√14/4). Vector AB is (0 - √2/4, 0 - √14/4) = (-√2/4, -√14/4). Vector AC is (3√2/2 - √2/4, 0 - √14/4) = (5√2/4, -√14/4). To check if AL bisects the angle between AB and AC, we can use the angle bisector formula in vectors. The angle bisector direction should be a linear combination of the unit vectors in the directions of AB and AC. But maybe it's easier to compute the angles. Let's compute the angle between AB and AL, and between AL and AC, and see if they are equal. First, compute the angle between AB and AL. The vectors are AB: (-√2/4, -√14/4) and AL: (√2/4, -√14/4). The dot product is (-√2/4)(√2/4) + (-√14/4)(-√14/4) = (-2/16) + (14/16) = (12/16) = 3/4. The magnitude of AB is sqrt( (-√2/4)^2 + (-√14/4)^2 ) = sqrt( (2/16) + (14/16) ) = sqrt(16/16) = 1. Which makes sense because AB is given as 1. The magnitude of AL is sqrt( (√2/4)^2 + (-√14/4)^2 ) = sqrt(2/16 + 14/16 ) = sqrt(16/16) = 1. Wait, AL is length 1? Wait, no. Wait, point A is at (√2/4, √14/4), point L is at (√2/2, 0). So, the distance AL is sqrt( (√2/2 - √2/4)^2 + (0 - √14/4)^2 ) = sqrt( (√2/4)^2 + (√14/4)^2 ) = sqrt( (2/16) + (14/16) ) = sqrt(16/16) = 1. So, AL is indeed length 1. So, the angle between AB and AL: cos(theta1) = (AB · AL) / (|AB||AL|) = (3/4)/(1*1) = 3/4. Therefore, theta1 = arccos(3/4). Similarly, compute the angle between AL and AC. Vector AC is (5√2/4, -√14/4). Vector AL is (√2/4, -√14/4). The dot product is (5√2/4)(√2/4) + (-√14/4)(-√14/4) = (10/16) + (14/16) = 24/16 = 3/2. Wait, but the dot product can't be larger than the product of the magnitudes. Wait, what's the magnitude of AC? AC is given as 2. The magnitude of AL is 1. So, the dot product is 3/2. Then, cos(theta2) = (3/2)/(2*1) = (3/2)/2 = 3/4. Therefore, theta2 = arccos(3/4). Therefore, both angles are equal, which confirms that AL is indeed the angle bisector. Therefore, the coordinates check out, the lengths check out, the angle bisector is correctly placed, and the altitude is correctly calculated. So, all steps are verified. Therefore, the lengths of the sides are AB = 1, AC = 2, and BC = (3√2)/2. Final AnswerThe lengths of the other sides are boxed{2} and boxed{dfrac{3sqrt{2}}{2}}.</think>Given triangle (ABC) with altitude (AH) and angle bisector (AL), the segments (BH), (HL), and (LC) are in the ratio (1:1:4). The length of side (AB) is 1. We need to find the lengths of sides (AC) and (BC).1. Segments on (BC): - Let (BH = x), (HL = x), and (LC = 4x). - Total length (BC = BH + HL + LC = x + x + 4x = 6x).2. Angle Bisector Theorem: - (AL) is the angle bisector, so (frac{BL}{LC} = frac{AB}{AC}). - (BL = BH + HL = x + x = 2x) and (LC = 4x), thus (frac{BL}{LC} = frac{2x}{4x} = frac{1}{2}). - Given (AB = 1), we have (frac{1}{AC} = frac{1}{2}), so (AC = 2).3. Using the Altitude (AH): - In right triangles (ABH) and (AHC): - (AB^2 = AH^2 + BH^2 Rightarrow 1^2 = AH^2 + x^2 Rightarrow AH^2 = 1 - x^2). - (AC^2 = AH^2 + HC^2 Rightarrow 2^2 = AH^2 + (5x)^2 Rightarrow 4 = AH^2 + 25x^2). - Substituting (AH^2 = 1 - x^2) into the second equation: - (4 = (1 - x^2) + 25x^2 Rightarrow 4 = 1 + 24x^2 Rightarrow 24x^2 = 3 Rightarrow x^2 = frac{1}{8} Rightarrow x = frac{sqrt{2}}{4}).4. Length of (BC): - (BC = 6x = 6 times frac{sqrt{2}}{4} = frac{3sqrt{2}}{2}).Thus, the lengths of the other sides are (boxed{2}) (for (AC)) and (boxed{dfrac{3sqrt{2}}{2}}) (for (BC)).]

🔑:1.2. Answer: A C=2, B C=3 / sqrt{2}. By the property of the angle bisector, we have A C: A B=L C: L B=2, from which A C=2 A B=2. Next, let B H=x, then H C=5 x. By the Pythagorean theorem for triangles A B H and A H C, we get 1^{2}-x^{2}= A H^{2}=2^{2}-25 x^{2}, 24 x^{2}=3, x=1 / 2 sqrt{2}, B C=3 / sqrt{2}.

🧠:[<think>Okay, so I have this problem about a sequence of positive integers. The first two elements are 1 and 2. The rule is that the sum of any two different elements can't be in the sequence. I need to prove that for any natural number k, the number of elements in the sequence less than k is at most k/3 + 2. Hmm. Let me try to wrap my head around this.First, let's understand the sequence. The first two terms are 1 and 2. Now, the next term can't be 1+2=3, because then 3 would be in the sequence, and 1+2=3 is a sum of two different elements. So 3 can't be in the sequence. So the next term after 2 has to be at least 4.Wait, let me check that. If 3 is in the sequence, then 1 and 2 sum to 3, which would violate the rule. So 3 can't be in the sequence. Therefore, the third term is 4. Then, the fourth term can't be 1+4=5 or 2+4=6. So 5 and 6 are excluded. So the next term after 4 would be 7? Let's see. But wait, 4 and 2 sum to 6, which is already excluded, but 4 and 1 sum to 5, which is also excluded. So next term after 4 is 7. Then, adding 7, we need to make sure that 1+7=8, 2+7=9, 4+7=11 aren't in the sequence. So next term after 7 is... let's see, the next integer not excluded. 5 is excluded, 6 is excluded, 7 is in, 8 is excluded because 1+7=8. 9 is excluded because 2+7=9. 10 is possible? Wait, 10 isn't the sum of any two previous elements. 1+2=3, 1+4=5, 1+7=8, 2+4=6, 2+7=9, 4+7=11. So 10 is not a sum of any two previous terms. So maybe 10 is next. Then 10 is added.Wait, but wait a minute. Let me check again. When adding a new term, do we have to make sure that the sum of any two different elements (including the new one) is not in the sequence. So when adding 7, we have to exclude 1+7=8, 2+7=9, 4+7=11. So 8,9,11 can't be in the sequence. Then when adding 10, we have to exclude 1+10=11, 2+10=12, 4+10=14, 7+10=17. So 11,12,14,17 can't be in the sequence. But 11 was already excluded when we added 7. So the next term after 10 would be... 13? Let's check. 13: 1+13=14, which is excluded by adding 10. 2+13=15, which isn't excluded yet. 4+13=17, excluded by 10. 7+13=20, 10+13=23. So 15,20,23 would need to be excluded. So 13 can be added. Then, adding 13, we must exclude 14,15,17,20,23. Then next term is 16? Let me check. 16: 1+16=17 (excluded), 2+16=18, 4+16=20 (excluded), 7+16=23 (excluded), 10+16=26, 13+16=29. So 18,26,29 are excluded. So 16 can be added. Then next term would be 19? Because 1+19=20 (excluded), 2+19=21, 4+19=23 (excluded), 7+19=26 (excluded), 10+19=29 (excluded), 13+19=32, 16+19=35. So 21,32,35 are excluded. So 19 can be added. Hmm. Interesting.So the sequence seems to be 1,2,4,7,10,13,16,19,... Each time, adding 3? Wait, 1, then 2 (difference 1), then 4 (difference 2), then 7 (difference 3), then 10 (difference 3), then 13 (difference 3), then 16 (difference 3), then 19 (difference 3)... So after 4, each term is previous +3. Wait, that's not quite. From 4 to 7 is +3, 7 to 10 is +3, 10 to 13 is +3, 13 to 16 is +3, 16 to 19 is +3. So starting from 4, each term is +3. So if that's the case, the sequence would be 1,2,4,7,10,13,16,19,22,25,... each term after 4 is previous +3. Is that the case? Let me check if this pattern holds.Suppose we continue this way. Let's try adding 22. Then, 1+22=23 (excluded by 10?), wait, 10+13=23 was excluded when we added 13? Wait, maybe not. Let's see. Wait, the exclusion happens when adding a new element. So when we add 19, we have to check sums of 19 with all previous elements. So 1+19=20 (excluded), 2+19=21 (excluded), 4+19=23 (excluded), 7+19=26 (excluded), 10+19=29 (excluded), 13+19=32 (excluded), 16+19=35 (excluded). So adding 19 excludes 20,21,23,26,29,32,35. So the next number would be 22. Then adding 22: 1+22=23 (already excluded), 2+22=24, 4+22=26 (excluded), 7+22=29 (excluded), 10+22=32 (excluded), 13+22=35 (excluded), 16+22=38, 19+22=41. So 24,38,41 are excluded. So 22 can be added. Then the next term is 25? Let's check. 1+25=26 (excluded), 2+25=27, 4+25=29 (excluded), 7+25=32 (excluded), 10+25=35 (excluded), 13+25=38 (excluded), 16+25=41 (excluded), 19+25=44, 22+25=47. So 27,44,47 are excluded. So 25 can be added. Then 28? Let me check. 1+28=29 (excluded), 2+28=30, 4+28=32 (excluded), 7+28=35 (excluded), 10+28=38 (excluded), 13+28=41 (excluded), 16+28=44 (excluded), 19+28=47 (excluded), 22+28=50, 25+28=53. So 30,50,53 are excluded. So 28 can be added. Wait, so now the pattern is 22,25,28,... each time adding 3. So seems like after 4, every term is +3. But let's check if there's a point where this breaks.Alternatively, maybe the sequence is constructed by starting with 1,2, and then each subsequent term is the smallest integer that doesn't form a sum with any two previous terms. So perhaps after 4, each term is 3 more than the previous. If that's the case, then the sequence is 1,2,4,7,10,13,16,19,22,... So each term after 4 is previous +3. Let's verify.If we follow this pattern, then the sequence is 1,2,4,7,10,13,... So the differences are 1,2,3,3,3,3,... So the third term is 4, which is 1+2+1? Not sure. Wait, maybe it's similar to the greedy algorithm for avoiding arithmetic progressions or something. Hmm.But perhaps we can model this sequence as a set S where S starts with 1,2, and each subsequent element is the smallest number greater than the previous one such that no two elements in S sum to another element in S. So similar to sum-free sets. Wait, sum-free sets are sets where no two elements sum to another element. But here, the condition is even stricter: the sum of any two different elements is not in the set. So exactly sum-free.So S is a sum-free set. Then, a classic result in combinatorics is that the maximum size of a sum-free subset of {1,2,...,n} is O(n). Wait, but here, the set is infinite. However, the problem is about the number of elements less than k. So we need to bound the number of elements less than k in this sequence by k/3 + 2.Wait, if the sequence after 4 is 7,10,13,16,... which is 3n +1? Let's see. Starting from 4:4,7,10,13,16,19,22,... Each term is 3 more than the previous. So 4 + 3(n-1) = 3n +1. Wait, 4 is the third term, so maybe 3n +1 for n >=1: 4,7,10,... Hmm. Wait, 3n +1 when n=1 is 4, n=2 is 7, n=3 is 10, etc. So the terms after the second term are of the form 3n +1. Then, starting from 4, the sequence is 4,7,10,13,... So the elements are 1,2,4,7,10,13,...Therefore, if we consider the elements less than k, let's see. The first two elements are 1 and 2. Then starting from 4, each subsequent element is 3 more. So the elements after 2 are 4,7,10,..., up to some number less than k. The number of terms in this arithmetic progression starting at 4, difference 3, less than k is floor((k -1 -4)/3) +1. Wait, maybe. Let me think.Suppose we have an arithmetic progression starting at a, with difference d. The number of terms less than k is floor((k -1 -a)/d) +1. So for a=4, d=3, the number of terms less than k is floor((k -1 -4)/3) +1 = floor((k -5)/3) +1. Which is equal to floor((k -5)/3 +1) = floor((k -5 +3)/3) = floor((k -2)/3). Hmm. Wait, maybe not. Let me check with an example. Let k=10. Then terms less than 10 are 4,7. So two terms. Using the formula: (10 -5)/3 = 5/3 ≈1.666, floor is 1, +1 gives 2. Correct. For k=13, terms less than 13 are 4,7,10. Three terms. (13 -5)/3=8/3≈2.666, floor is 2, +1=3. Correct. So the formula works. So number of terms after 2 (i.e., starting from 4) less than k is floor((k -5)/3) +1 = floor((k -2)/3). Wait, (k -5)/3 +1 = (k -5 +3)/3 = (k -2)/3. So floor((k -2)/3). So total number of elements less than k would be 2 (for 1 and 2) plus floor((k -2)/3). So total is 2 + floor((k -2)/3). Now, we need to show that the number of elements less than k is at most k/3 +2.Note that floor((k -2)/3) ≤ (k -2)/3. So 2 + floor((k -2)/3) ≤ 2 + (k -2)/3 = (6 +k -2)/3 = (k +4)/3. Hmm. Wait, but the problem statement says it should be at most k/3 +2. But (k +4)/3 = k/3 + 4/3. Which is less than k/3 +2 when 4/3 <2, which is true. So actually, the bound we get here is even better: (k +4)/3, which is less than k/3 +2. Because (k +4)/3 = k/3 +4/3 < k/3 +2. So that would satisfy the required bound. Wait, but maybe my assumption is wrong. Because in the example earlier, with k=10, the number of elements less than 10 is 1,2,4,7. That's four elements. According to the formula 2 + floor((10 -2)/3) = 2 + floor(8/3)=2+2=4. Correct. Then according to the upper bound k/3 +2, which is 10/3 +2 ≈3.333 +2=5.333. So 4 ≤5.333. So that's okay. The upper bound is not tight. But according to the previous calculation, our actual number is (k +4)/3, which is less than k/3 +2. So maybe the problem is stated correctly, and perhaps my initial reasoning is correct, but the key is to formalize it.But wait, maybe the sequence isn't exactly the arithmetic progression starting at 4 with difference 3. Because when we added 4,7,10,13,... but maybe at some point, we can't just add 3 each time? For example, when adding 22, 25, 28, etc., maybe some number in between would become allowable. But in my earlier reasoning, when adding each term, we exclude the next few numbers, but since we're adding numbers spaced 3 apart, their sums with previous numbers would block numbers that are spaced 1, 2, 3, etc., but given that each term is 3 apart, the sums would cover numbers spaced 1+3=4, 2+3=5, etc., but perhaps the key is that by spacing them 3 apart, the forbidden sums are spaced in a way that allows the next number to be 3 more. However, this is just heuristic. Maybe I need a more formal approach.Alternatively, suppose that the sequence S contains 1,2, and all subsequent numbers are congruent to 1 modulo 3. Because 4 is 1 mod 3 (4=3*1 +1), 7=3*2 +1, 10=3*3 +1, etc. So the terms after 2 are all numbers ≡1 mod3. Let's check if that's the case. Then, any two numbers ≡1 mod3 would sum to ≡2 mod3, which is not in the sequence (since after 2, all elements are ≡1 mod3). The sum of 1 (≡1 mod3) and any element ≡1 mod3 would be ≡2 mod3, which is not in the sequence. The sum of 2 (≡2 mod3) and any element ≡1 mod3 would be ≡0 mod3, which is not in the sequence (since 0 mod3 is not in S except for 1,2, but 0 mod3 is 3,6,9,... which are excluded). So if S consists of 1,2, and all numbers ≡1 mod3 starting from 4, then indeed, the sum of any two elements would be ≡2,0, or1 mod3. But elements in S are 1,2, and 1 mod3. So the sums:1 +1 mod3=2 mod3 (not in S except for 2, but 1 +1=2 is allowed? Wait, but the problem states "the sum of any two different elements". Wait, the problem says "the sum of any two different elements of the sequence is not an element of the sequence". So even if two different elements sum to an element, it's not allowed. So if we have 1 and 1, but the elements are positive integers, and the sequence elements are distinct. So the sequence has distinct elements. So 1 is only once. So the sum of 1 and 2 is 3, which is not in the sequence. The sum of 1 and 4 is 5, not in the sequence. The sum of 2 and4 is6, not in the sequence. The sum of 1 and7 is8, not in the sequence. The sum of 2 and7 is9, not in the sequence. The sum of4 and7 is11, not in the sequence. So indeed, if we take 1,2, and all numbers ≡1 mod3 starting from4, then the sum of any two different elements is not in the sequence.Therefore, such a sequence satisfies the conditions. Now, in this case, the number of elements less than k would be:- 1 and2: two elements.- Then, numbers ≡1 mod3 starting from4 up to less than k. The numbers are4,7,10,...,m where m <k and m≡1 mod3. The count of such numbers is floor((k -1 -1)/3) = floor((k -2)/3). Wait, the first term is4, which is1 mod3, so the numbers are4=3*1 +1,7=3*2 +1,..., up to the largest less thank. The number of terms is floor((k -4)/3) +1. For example, ifk=10, then (10 -4)/3=2, floor(2)=2, +1=3 terms:4,7,10. But wait, 10 is not less than10. So actually, we need numbers less than k. So the largest term is the largest number ≡1 mod3 less thank. Ifk ≡1 mod3, then it'sk-1. Otherwise, it'sk - ( (k mod3) -1 ) ifk mod3 ≥1. Hmm, maybe another approach.The numbers ≡1 mod3 starting from4 are4,7,10,..., which can be written as3n +1 forn ≥1. So we have3n +1 <k ⇒n <(k -1)/3. Thus, the number of terms is floor((k -1)/3 -1) = floor((k -4)/3). Wait, let's see:Ifk=7: 3n +1 <7 ⇒3n <6 ⇒n <2 ⇒n=1, so 4. So one term. floor((7-4)/3)=1. Correct.Ifk=10:3n +1 <10 ⇒3n <9 ⇒n <3 ⇒n=1,2: 4,7. Two terms. floor((10-4)/3)=2. Correct.Wait, but (k -4)/3. Ifk=7: (7-4)/3=1, floor(1)=1. Correct.Ifk=10: (10-4)/3=2, floor=2. Correct.Ifk=13: (13-4)/3=3, floor=3. Then numbers4,7,10,13. But13 is not less than13. So the count should be3, which is floor((13-4)/3)=3. But13-4=9, 9/3=3. So floor(3)=3. But the numbers less than13 are4,7,10. Three numbers. So correct.Thus, the number of terms≡1 mod3 less thank isfloor((k -4)/3). Then, adding the two initial terms1 and2, the total number is2 +floor((k -4)/3). Now, we need to show that2 +floor((k -4)/3) ≤k/3 +2. Sincefloor((k -4)/3) ≤(k -4)/3, then2 +floor((k -4)/3) ≤2 + (k -4)/3 = (6 +k -4)/3 = (k +2)/3. But(k +2)/3 =k/3 +2/3, which is less thank/3 +2. So(k +2)/3 ≤k/3 +2. Therefore, the total number is≤k/3 +2. Hence proved.Wait, but the problem states "the number of elements of the sequence that are less than k is at mostk/3 +2". According to our construction, the number is2 +floor((k -4)/3) ≤(k +2)/3. Since(k +2)/3 ≤k/3 +2 (because2/3 ≤2), then yes. So the required inequality holds.But wait, in the problem statement, the upper bound isk/3 +2, which is larger than our upper bound of(k +2)/3. Therefore, since(k +2)/3 ≤k/3 +2, our sequence's count is within the required bound.Therefore, if the sequence is constructed as1,2, and all numbers≡1 mod3 starting from4, then the number of elements less thank is2 +floor((k -4)/3), which is≤k/3 +2.But the problem states that "the elements of a sequence are positive integers, the first two elements are1 and2. The sum of any two different elements of the sequence is not an element of the sequence." So we need to prove that for any such sequence (not just the specific one we constructed), the number of elements less thank is at mostk/3 +2.Ah, so the problem isn't about a specific sequence, but about any sequence satisfying the given conditions. So our previous reasoning assumed a specific sequence, but the problem requires that for any possible sequence meeting the criteria, the bound holds. Therefore, we need a more general approach.So how can we approach this? Let's consider that S is a sequence starting with1,2, and it's sum-free (i.e., no two distinct elements sum to another element). We need to show that the number of elements in S less than k is at mostk/3 +2.Sum-free sets have been studied. One classic result is that any sum-free subset of {1,2,...,n} has size at most⌈n/2⌉. But here, we're dealing with an infinite set, and the bound is different. The required bound is linear in k with coefficient1/3.Another classic result is that the set of numbers in {1,2,...,n} that are congruent to some value modulo3 (for example, all numbers congruent to2 mod3) form a sum-free set. The size of such a set is roughlyn/3, which matches the bound here.Therefore, perhaps the maximum size is achieved by such a set. In our earlier construction, we had a set with size roughlyk/3 +2. But the problem allows for any sum-free set starting with1 and2. So maybe the maximum possible size is indeed achieved by the set {1,2} union with numbers≡1 mod3 starting from4. Then, the size is≈k/3 +2. So any other sum-free set containing1 and2 cannot have more elements.Therefore, to prove the bound, we need to show that any such sequence cannot have more thank/3 +2 elements less thank.Let me think of another approach. Suppose S is a sum-free set containing1 and2. Let A = S ∩ [1, k). We need to show |A| ≤k/3 +2.Since S is sum-free, for any distinct a,b ∈ S, a +b ∉ S. In particular, since1 and2 are in S, their sum3 is not in S. Also,1 and any other elementc ∈ S,1 +c ∉ S. Similarly,2 and any other elementd ∈ S,2 +d ∉ S.Therefore, if we consider the elements of S beyond2, sayc_1,c_2,c_3,..., then eachc_i must satisfy:1 +c_i ∉ S,2 +c_i ∉ S,and also for any two elementsc_i,c_j (i ≠j),c_i +c_j ∉ S.This seems complex. Maybe we can partition the numbers into residue classes mod3. Let's consider numbers modulo3.Numbers can be0,1,2 mod3. Let's see:If a number is0 mod3, say3m, then adding1 to it gives3m +1 mod3, which is1 mod3. Adding2 to it gives3m +2 mod3, which is2 mod3.If a number is1 mod3, say3m +1, adding1 gives3m +2 mod3, adding2 gives3m +3=3(m +1)≡0 mod3.If a number is2 mod3, say3m +2, adding1 gives3m +3≡0 mod3, adding2 gives3m +4≡1 mod3.Therefore, if we choose numbers in certain residue classes, we can avoid sums.For example, if we take numbers≡1 mod3, then:1 + (3m +1)=3m +2≡2 mod3,2 + (3m +1)=3m +3≡0 mod3,So sums with1 and2 will be in0 or2 mod3. If S contains only1,2, and numbers≡1 mod3, then the sums1 +c_i≡2 or0 mod3, which are not in S (except for1 and2). But we have to ensure that sums between the numbers≡1 mod3 are not in S. Ifc_i andc_j are≡1 mod3, thenc_i +c_j ≡2 mod3, which is not in S (except for2). So sincec_i,c_j ≥4, their sum is at least8, which is greater than2, so2 is not in the sum set. Hence, the set {1,2} ∪ {numbers≡1 mod3 starting from4} is sum-free.Similarly, another possible choice could be numbers≡2 mod3. Let's check:If we have numbers≡2 mod3, then1 + (3m +2)=3m +3≡0 mod3,2 + (3m +2)=3m +4≡1 mod3.So sums with1 and2 would be0 or1 mod3. If S contains1,2, and numbers≡2 mod3 starting from5 (since2 is already in S), then the sums1 +c_i≡0 mod3 (not in S except for numbers≡0 mod3, which are excluded), and2 +c_i≡1 mod3 (not in S except for1). But1 is already in S, but2 +c_i would be at least2 +5=7, which is greater than1, so1 is not in the sum set. Also, sums between the numbers≡2 mod3: (3m +2)+(3n +2)=3(m +n +1) +1≡1 mod3. So the sum is≡1 mod3. If S doesn't contain numbers≡1 mod3 (except1), then these sums are not in S. But since we already have1 in S, we need to make sure that1 is the only element≡1 mod3. Therefore, if we take S as{1,2} ∪ {numbers≡2 mod3 starting from5}, then this is also sum-free. However, this set would have numbers2,5,8,11,... So similar to the previous case, but with different residues.However, the size would be similar: numbers≡2 mod3 starting from5: the count less thank would be floor((k -5)/3) +1, plus the two elements1 and2. So total is2 +floor((k -5)/3) +1=3 +floor((k -5)/3). Which is similar to the previous case, but slightly different. For example, ifk=10, numbers≡2 mod3 starting from5 are5,8. So count is2. Then total elements less than10 are1,2,5,8:4. Which is same as the previous case. So floor((10 -5)/3)=1, +1=2. 3 +1=4. Similarly, the previous case gave2 +floor((10 -4)/3)=2 +2=4. So same count.So regardless of the residue class chosen (1 or2 mod3), the count is similar. Therefore, the maximum size is achieved by such constructions, giving the bound ofk/3 +2.However, the problem states that the sequence starts with1 and2, and we need to show that any such sequence has at mostk/3 +2 elements less thank. So even if someone tries to include more numbers by choosing a different residue class or a different strategy, they can't exceed this bound.To formalize this, let's consider the following:Let S be a sum-free set containing1 and2. Let A = S ∩ [1, k). We need to show |A| ≤k/3 +2.Let’s partition the numbers in A into three categories:1. The elements1 and2.2. The elements greater than2.For the elements greater than2, say they are in the setB. ThenB must be sum-free, and also, for anyx ∈ B,1 +x ∉ S and2 +x ∉ S. Since S is sum-free,1 +x and2 +x cannot be in S.But B is a subset ofS, so1 +x and2 +x ∉ S. However, since S may include other elements, we need to ensure that these sums are not in S. But B itself is a subset of S, so for anyx,y ∈ B, x +y ∉ S. However, S also includes1 and2. So it's a bit more complex.Alternatively, perhaps we can model this after the concept of sum-free sets in additive number theory. A standard approach to bound the size of sum-free sets is using the theorem by Erdős, which might not be directly applicable here. However, a simple method is the following:Consider that for anyx ∈ B, the numbersx +1 andx +2 are not in S. Because ifx +1 were in S, then1 +x would be in S, which is forbidden. Similarly, ifx +2 were in S, then2 +x would be in S, which is forbidden. Therefore, for eachx ∈ B, bothx +1 andx +2 are excluded from S.Additionally, for any two distinct elementsx,y ∈ B, their sumx +y is not in S.Therefore, each elementx ∈ B excludes:- x +1 andx +2 from S,- and for everyy ∈ B,y >x, x +y is excluded from S.But the key exclusions arex +1 andx +2. Because these exclusions apply to all elementsx ∈ B, we can model the available numbers as being spaced out such that eachx ∈ B requires that the next two numbers are excluded.If we think of the number line, each elementx ∈ B blocks the positionsx +1 andx +2. Therefore, the elements ofB must be spaced such that between any two elements ofB, there are at least two numbers that are excluded. This is similar to placing objects with a minimum distance between them.But actually, more precisely, for eachx ∈ B, x +1 andx +2 are excluded, so the next possible element afterx would have to be at leastx +3. Thus, the elements ofB must be at least3 apart. Therefore, the setB is a subset of an arithmetic progression with difference at least3. However, they don't have to be exactly3 apart, but they can't be closer than3.Therefore, the maximum number of elements inB less thank is roughlyk/3. Since each element needs at least3 numbers: itself and the next two excluded. Therefore, the maximum size ofB is floor((k -3)/3) +1, which is≈k/3.Adding the two elements1 and2, the total size is≈k/3 +2, which matches the required bound.To formalize this, supposeB hasm elements. Each elementx_i ∈ B requires thatx_i +1 andx_i +2 are not in S. Therefore, the intervals [x_i, x_i +2] for eachx_i ∈ B are forbidden for other elements. If the elements ofB are orderedx_1 <x_2 <... <x_m, thenx_{i+1} ≥x_i +3. Therefore, the sequencex_1,x_2,...,x_m forms an increasing sequence withx_{i+1} ≥x_i +3. Therefore, the minimal possiblex_i is3i -2 + starting point. Wait, if starting from3, but in our case, the first element ofB is4 (since1 and2 are already in S, and3 is excluded because1 +2=3). So the first element ofB is4.Therefore, starting from4, each subsequent element is at least3 more than the previous. Therefore, the elements ofB are≥4,7,10,..., forming an arithmetic progression with difference at least3. Hence, the number of elements inB less thank is≤ floor((k -4)/3) +1. Therefore, the total number of elements inS less thank is2 +floor((k -4)/3) +1=3 +floor((k -4)/3). Wait, no. Wait,B is the elements greater than2, so starting from4. So number of elements inB less thank isfloor((k -4)/3) +1. Thus, total elements inS less thank are2 +floor((k -4)/3) +1=3 +floor((k -4)/3). Wait, that seems conflicting with earlier examples. Wait, let me check.Ifk=10, elements inB less than10 are4,7. So two elements. floor((10-4)/3)=2, so floor((k -4)/3)=2. Then total elements=2 +2=4. Which matches. The formula would be2 +floor((k -4)/3). So yes, the elements inB less thank arefloor((k -4)/3) +1. Wait, wait:If we start from4, the number of terms less thank in the arithmetic progression4,7,10,... isfloor((k -1 -4)/3) +1=floor((k -5)/3) +1. For example,k=10:floor((10-5)/3)=1, +1=2. Correct (4,7). Fork=13:floor((13-5)/3)=2, +1=3. Correct (4,7,10). So yes, number of elements inB less thank isfloor((k -5)/3) +1=floor((k -5)/3 +1)=floor((k -2)/3). Therefore, total elements inS less thank is2 +floor((k -2)/3). Wait, this contradicts previous statements. Let me clarify:The elements inB are4,7,10,..., which is an arithmetic progression starting at4, difference3. The number of terms less thank is given by the largest integern such that4 +3(n -1) <k. So:4 +3(n -1) <k ⇒3n -3 +4 <k ⇒3n +1 <k ⇒3n <k -1 ⇒n <(k -1)/3 ⇒n= floor((k -1)/3 -1)=floor((k -4)/3).Wait, examplek=10:floor((10 -4)/3)=2. So two terms:4,7. Correct.k=13:floor((13-4)/3)=3. Three terms:4,7,10. Correct.So number of terms inB less thank isfloor((k -4)/3). Therefore, total elements inS less thank is2 +floor((k -4)/3). Now, we need to show that2 +floor((k -4)/3) ≤k/3 +2.Sincefloor((k -4)/3) ≤(k -4)/3, then2 +floor((k -4)/3) ≤2 + (k -4)/3 = (6 +k -4)/3 = (k +2)/3. Now,(k +2)/3 ≤k/3 +2 because(k +2)/3 =k/3 +2/3 ≤k/3 +2. Therefore, the total number of elements is≤k/3 +2, as required.Hence, proved.Final AnswerThe maximum number of elements less than ( k ) is at most ( boxed{frac{k}{3} + 2} ).</think>Given a sequence of positive integers starting with 1 and 2, where the sum of any two different elements is not an element of the sequence, we need to prove that the number of elements less than any natural number ( k ) is at most ( frac{k}{3} + 2 ).1. Understanding the Sequence: - The sequence starts with 1 and 2. - The sum of any two different elements cannot be in the sequence. - After 1 and 2, the next elements must be chosen such that their sum with any previous element is not in the sequence.2. Constructing the Sequence: - The third element must be 4 (since 3 is excluded as 1 + 2). - Subsequent elements are chosen to be the smallest integers that do not form a sum with any previous elements. This results in elements 4, 7, 10, 13, 16, 19, ..., which form an arithmetic progression starting from 4 with a common difference of 3.3. Residue Class Analysis: - Elements after 2 are congruent to 1 modulo 3 (i.e., 4, 7, 10, ...). - Sums of elements in this sequence do not fall into the same residue class, ensuring the sequence remains sum-free.4. Counting Elements Less Than ( k ): - The first two elements are 1 and 2. - The number of elements in the arithmetic progression 4, 7, 10, ... less than ( k ) is given by ( leftlfloor frac{k - 4}{3} rightrfloor ).5. Bounding the Count: - Total elements less than ( k ) include 1, 2, and the elements in the arithmetic progression. - The number of elements in the progression is ( leftlfloor frac{k - 4}{3} rightrfloor ). - The total count is ( 2 + leftlfloor frac{k - 4}{3} rightrfloor ).6. Final Bound: - Using the inequality ( leftlfloor frac{k - 4}{3} rightrfloor leq frac{k - 4}{3} ), we get: [ 2 + leftlfloor frac{k - 4}{3} rightrfloor leq 2 + frac{k - 4}{3} = frac{k + 2}{3} ] - Since ( frac{k + 2}{3} leq frac{k}{3} + 2 ), the bound is established.Thus, the number of elements less than ( k ) in the sequence is at most ( boxed{frac{k}{3} + 2} ).,<think>Okay, so I have this problem here about a sequence of positive integers. The first two elements are 1 and 2. The rule is that the sum of any two different elements can't be in the sequence. I need to prove that for any natural number k, the number of elements less than k is at most k/3 + 2. Hmm, let's see how to approach this.First, let's make sure I understand the problem correctly. The sequence starts with 1, 2. Then every subsequent element must be chosen such that if you take any two different elements and add them, the result isn't in the sequence. So, for example, the third element can't be 1+2=3. Then the next element after that also has to avoid sums of any two previous elements. So, the sequence is being built in a way that avoids having any element that's the sum of two others. Interesting.So, the challenge is to show that such a sequence can't have too many elements below any given k. Specifically, the number of elements less than k can't exceed k/3 + 2. That seems like a linear bound, which makes me think that maybe the sequence has to be spaced out in some regular way, perhaps with elements at least 3 apart? But 1 and 2 are consecutive, so that might not be the case. Maybe there's a pattern or a structure here that ensures the elements are not too dense.Let me try constructing the sequence step by step to see what's happening. Starting with 1 and 2. The next number can't be 3 (since 1+2=3), so the next possible number is 4. Now, adding 4 to the previous elements: 1+4=5, 2+4=6. So 5 and 6 can't be in the sequence. The next number after 4 would have to be 7. Then, 1+7=8, 2+7=9, 4+7=11, so 8, 9, 11 are excluded. Next number could be 10? Wait, but 10 isn't the sum of any two previous elements. Let's check: 1+10=11 (which is excluded), 2+10=12, 4+10=14. Since 11 is already excluded, 10 doesn't create a sum that's already in the sequence. Wait, but the rule is that the sum of any two different elements is not in the sequence. So, if 10 is added, we need to ensure that no two elements (including 10) sum to another element. So, if we add 10, then we have to check if 1+10=11 is in the sequence. If 11 isn't there, then it's okay. Similarly, 2+10=12, which isn't in the sequence yet, and 4+10=14, which also isn't there. So 10 is allowed. Wait, but then maybe the next number after 7 is 10? But maybe there's a smaller number that can be added. Let's see.After 1,2,4,7, let's see what's next. The next number after 7 would have to be the smallest number that isn't excluded by the previous elements. Let's check 5: 1+4=5, which is already excluded. 6 is excluded because 2+4=6. 7 is already in the sequence. 8 is excluded by 1+7=8. 9 is excluded by 2+7=9. 10: as above, 1+10=11 (not in the sequence), 2+10=12 (not in), 4+10=14 (not in), 7+10=17 (not in). So 10 is okay. Wait, but if we add 10, then any pair including 10 must not sum to an existing element. Since none of the sums 1+10, 2+10, 4+10, 7+10 are in the sequence, 10 is allowed. So, maybe the sequence goes 1, 2, 4, 7, 10, 13,... Each time adding 3? Wait, 1 to 2 is +1, 2 to 4 is +2, 4 to 7 is +3, 7 to 10 is +3, 10 to 13 is +3. Hmm, is that a pattern?Wait, but if we have 1,2,4,7,10,13,... then each step after 4 is adding 3. But let's check why 7 was chosen. After 1,2,4, the next number can't be 5 (1+4), 6 (2+4), 7: is 7 allowed? Let's check sums. 1+2=3 (not in the sequence), 1+4=5 (not in), 2+4=6 (not in). So 7 is okay. Then next number after 7: 8 (1+7), 9 (2+7), 10 (not a sum of any two previous elements). So 10 is okay. Then 11 (1+10), 12 (2+10), 13 (4+9 but 9 isn't in the sequence, so 13 is okay). Wait, 4+10=14, so 14 is excluded. So 13 is allowed. Then 13: check sums with previous elements. 1+13=14 (excluded), 2+13=15 (excluded?), 4+13=17 (excluded?), 7+13=20 (excluded?), 10+13=23 (excluded). So 13 is okay.So the sequence is 1,2,4,7,10,13,... each time adding 3, starting from 4? So 1,2, then 4, then +3 each time? That seems to be a possible way to construct the sequence. But maybe there's another way? For example, after 1,2,4,7,10,13,... but maybe inserting other numbers in between?Wait, let's test that. Suppose after 1,2,4,7, instead of adding 10, could we add 8? But 1+7=8, so 8 is excluded. 9 is excluded because 2+7=9. 10 is allowed. So, no, 8 and 9 are excluded. Similarly, 11 is excluded by 1+10=11 if we add 10. So next is 10, then 13, etc. So it seems that once we get past 4, the sequence progresses by adding 3 each time. So the sequence is 1,2,4,7,10,13,16,19,... each term after 4 is 3 more than the previous. If that's the case, then the number of elements less than k would be roughly k/3. Because starting from 4, each term is spaced 3 apart. But the first few terms are 1,2,4, so maybe the total number is (k - 1)/3 + some constant.But the problem states that the number of elements less than k is at most k/3 + 2. So if our constructed sequence is 1,2,4,7,10,... then up to k, say k=10, the elements less than 10 are 1,2,4,7, which is 4 elements. According to the formula, k/3 +2 = 10/3 +2 ≈ 3.33 +2 = 5.33, so 4 is less than that. If k=13, elements less than 13 are 1,2,4,7,10: 5 elements. Formula gives 13/3 +2 ≈4.33 +2≈6.33, so 5 is less. So this sequence seems to fit within the bound.But is this the maximal possible sequence? Or can there be other sequences with more elements still satisfying the condition? The problem says "for any natural number k, the number of elements of the sequence that are less than k is at most k/3 +2". So regardless of how the sequence is constructed (as long as it follows the given rules), the number of elements below k can't exceed k/3 +2. So even if someone tries to pack in as many elements as possible, they can't exceed that bound.So, to prove this, maybe we can use induction or some kind of counting argument. Let me think.First, let's note that the sequence must avoid containing any sum of two distinct elements. So, for any two elements a and b (a ≠ b), a + b is not in the sequence. So, the sequence is a kind of "sum-free set" over the positive integers. However, in typical sum-free sets, we often consider sets where no two elements sum to another element. So this is similar.In additive number theory, a sum-free set is a set of numbers such that no two elements add up to another element. The classic example is the set of all odd numbers, or the upper half of the integers. However, in our case, the sequence must start with 1 and 2, which are not both odd, so the usual examples might not apply here.But there are known results about the maximum size of a sum-free subset of {1, 2, ..., n}. For example, the maximum size is roughly n/2, by taking the upper half. But our problem is different because it's an infinite sequence, and we have a specific starting pair. Also, the bound here is k/3 +2, which is much smaller than n/2. So maybe the structure here is more restrictive.Alternatively, maybe we can model this sequence as similar to a greedy sequence avoiding sums. Let's see. Starting with 1 and 2, then each subsequent term is the smallest possible number that doesn't form a sum with any two previous terms.But wait, in the example I constructed, after 1,2,4,7,10,13,... each term is 3 more than the previous starting from 4. So 4,7,10,13,... So that's an arithmetic progression with difference 3. Maybe this is the key structure. If the sequence is built in such a way that after a certain point, the terms are in arithmetic progression with difference 3, then the density would be 1/3, leading to the k/3 term in the bound. The +2 would account for the initial terms that don't fit into the progression.But how can we formalize this? Let's think about the general structure. Suppose that after some initial terms, the sequence continues with an arithmetic progression of difference 3. Let's see why that might be necessary.Suppose we have a term a_n. The next term a_{n+1} must be chosen such that a_{n+1} + a_i is not in the sequence for any i < n+1. If we set a_{n+1} = a_n + 3, then we need to check that (a_n + 3) + a_i is not in the sequence. If the previous terms are spaced at least 3 apart, maybe this can be maintained. But I need to check.Alternatively, maybe we can show that each new term after 4 must be at least 3 more than the previous term. If we can establish that a_{n+1} ≥ a_n +3, then the number of terms less than k would be at most (k -1)/3 + c, where c is a constant for the initial terms. This would lead to the bound given.So, let's try to show that each term is at least 3 more than the previous one after some point. Let's suppose we have a sequence starting with 1,2,4. Then 4 is 2 more than 2. Then 7 is 3 more than 4. Then 10 is 3 more than 7, etc. So after 4, each term is 3 more. But why can't we have a smaller difference?Suppose we have a term x, and we want to choose the next term y. y must not be such that y + any previous term is in the sequence. Also, y itself can't be the sum of any two previous terms.Wait, but actually, the condition is that the sum of any two different elements is not in the sequence. So, the sequence must not contain any element that is the sum of two different elements. Therefore, when adding a new element y, we need to ensure two things:1. y is not equal to any sum of two previous elements.2. For all previous elements a_i, the sum a_i + y is not in the sequence.But actually, the problem states "the sum of any two different elements of the sequence is not an element of the sequence". So, in other words, for any two distinct elements a and b, a + b is not in the sequence. Therefore, when adding a new element y, we must ensure that:- For all existing elements a_i, a_i + y is not in the sequence.Additionally, y itself should not be a sum of two existing elements.Therefore, when choosing y, it must not be expressible as a sum of two previous elements, and also, adding y to any previous element must not result in an element that's already in the sequence.So, in constructing the sequence, after 1 and 2, the next element cannot be 3 (since 1+2=3), so we choose 4. Now, 4 is not a sum of 1 and 2 (1+2=3, 1+1=2, 2+2=4, but elements must be distinct, so 2+2 is not allowed since they are the same element). Wait, the problem says "any two different elements", so duplicates are allowed? Wait, no, the elements are positive integers, so they are distinct. Wait, the problem says "elements are positive integers", but it doesn't specify they are distinct. Wait, but sequences can have repeated elements unless specified otherwise. Wait, but in the problem statement, it's not mentioned whether the sequence has distinct elements. Wait, but the first two elements are 1 and 2, which are distinct. But if the sequence can have repeated elements, then the problem becomes different. Wait, but the standard definition of a sequence allows repeated elements unless stated otherwise. But in the context of problems like this, usually, the sequence is of distinct elements. But the problem statement doesn't specify. However, the problem says "the sum of any two different elements of the sequence is not an element of the sequence". So even if the sequence had repeated elements, the sum of two different elements (i.e., two distinct positions) would not be in the sequence. However, if the same number appears multiple times, then adding two instances of that number would be a sum of two different elements (since they are in different positions), so that sum would be 2 times the number, which would have to not be in the sequence. But since the elements are positive integers, 2 times a number is another number, which might or might not be in the sequence.However, given that the problem is about the number of elements less than k, it's more natural to consider that the sequence consists of distinct elements. Otherwise, you could have infinitely many copies of 1 and 2, but then the number of elements less than k would be unbounded, which contradicts the bound given. Therefore, the sequence must consist of distinct positive integers. So, we can assume that all elements are distinct.Therefore, the problem is about a sequence of distinct positive integers starting with 1 and 2, such that no two distinct elements sum to another element. We need to show that the number of elements less than k is at most k/3 + 2.Okay, that makes more sense. So, the sequence is a set of distinct positive integers, starting with 1 and 2, and it's sum-free (no two distinct elements sum to another element). We need to prove the density is at most k/3 + 2.In additive combinatorics, a classic result is that any sum-free subset of {1, 2, ..., n} has size at most ⌈n/2⌉. But in our case, the set is infinite, and we have a specific starting pair. The bound here is much stricter (k/3 vs. k/2), so the structure must be more specialized.Perhaps the key is that once you have 1 and 2, the next elements have to be spaced out to avoid creating sums that would conflict with existing elements. Let's try to formalize this.Let the sequence be a_1, a_2, a_3, ..., where a_1=1, a_2=2, and each subsequent a_n is the smallest integer greater than a_{n-1} such that no two distinct elements sum to another element.We need to show that for any k, the number of a_i < k is ≤ k/3 + 2.Perhaps we can partition the positive integers into residue classes modulo 3 and show that the sequence can contain at most one number from each residue class, which would imply the density is 1/3. But 1 and 2 are in different residue classes (1 mod 3 and 2 mod 3), so maybe after that, all subsequent numbers must be 0 mod 3. Let's check.If we have 1 and 2, then adding a number congruent to 0 mod 3: say 3. But 1+2=3, so 3 is excluded. Next, 6: 1+5=6, but 5 isn't in the sequence yet. Wait, this approach might not work. Alternatively, maybe after 1 and 2, the sequence can only include numbers congruent to 0 mod 3 starting from 3, but 3 is excluded. Then 6, but 1+5=6, but 5 isn't in the sequence. So maybe 6 is allowed if 5 isn't present. Hmm, perhaps not.Alternatively, looking at the constructed sequence: 1, 2, 4, 7, 10, 13, 16,... These numbers modulo 3 are 1, 2, 1, 1, 1, 1, 1,... Wait, no. 1 mod 3, 2 mod 3, 1 mod 3 (4=3+1), 7=6+1 mod 3=1, 10=9+1 mod3=1, etc. So starting from 4, they are all 1 mod 3. But then, adding these numbers, their sums would be 1+1=2 mod3, 1+2=0 mod3, etc. But maybe this structure avoids creating sums within the sequence.Wait, let's see. If after 4, all elements are 1 mod3, then:- 1 (mod3) + 2 (mod3) = 0 mod3, which isn't in the sequence (since after 4, all elements are 1 mod3, and 0 mod3 numbers like 3,6,9 are excluded).- 1 (mod3) + 1 (mod3) = 2 mod3, which is already occupied by 2, so adding another 1 mod3 element would create a sum (1 + 1 mod3 element) = 2 mod3, but 2 is already in the sequence. Wait, 1 + 4 = 5, which is 2 mod3. Since 2 is in the sequence, is 5 excluded? Wait, the rule is that the sum of any two different elements is not in the sequence. So, 1 + 4 = 5. If 5 were in the sequence, that would be bad. But 5 isn't in the sequence, so it's okay. So even though 5 is 2 mod3, and 2 is in the sequence, since 5 isn't actually in the sequence, it's allowed.Therefore, having multiple 1 mod3 elements doesn't necessarily conflict because their sums could land in residue classes that aren't present in the sequence. So maybe this approach works. But how does this help us with the bound?Alternatively, perhaps considering that once we have 1 and 2, the sequence must avoid numbers that are 1 + previous elements or 2 + previous elements. So every time we add a number x, we have to exclude all numbers of the form x + 1 and x + 2 from the sequence. Therefore, each new element x effectively blocks x +1 and x +2 from being in the sequence. If the sequence is constructed in a way that these blocked intervals don't overlap, then the minimal spacing would be 3. Hence, the density is 1/3.But let's try to formalize this idea. Suppose we have elements a_1=1, a_2=2. Then, when we add a_3=4, we block 4+1=5 and 4+2=6. Then, when we add a_4=7, we block 8 and 9. Then a_5=10 blocks 11 and 12, and so on. Each new element blocks the next two numbers. Therefore, the spacing between elements after 4 is 3: 4,7,10,13,... each 3 apart. Therefore, the number of elements less than k in this part of the sequence is roughly (k -4)/3. Then adding the initial elements 1,2,4 gives a total of approximately (k -4)/3 +3. Which is k/3 -4/3 +3 = k/3 + 5/3 ≈ k/3 +1.666, which is within the bound of k/3 +2. So this example sequence fits the bound.But to prove that any such sequence must satisfy this bound, we need to show that no matter how you construct the sequence (following the given rules), you can't have more than k/3 +2 elements below k.Perhaps we can model the sequence as a set S containing 1 and 2, and such that for any x, y ∈ S with x ≠ y, x + y ∉ S. We need to show that |S ∩ [1, k)| ≤ k/3 +2.One approach is to consider that each element x > 2 in S "blocks" two numbers: x +1 and x +2, because if x is in S, then 1 + x and 2 + x cannot be in S. Therefore, each such x prevents at least two numbers from being in S. Additionally, the elements themselves take up space. So maybe we can use an amortized analysis, counting the number of elements and the number of blocked numbers.Let me try this. Suppose S is our sequence. Let’s divide the positive integers into intervals or blocks and analyze how many elements can be in each block.Alternatively, think of each element x ≥3 as being responsible for excluding the numbers x +1 and x +2. Therefore, for each such x, we have three numbers: x, x+1, x+2, where only x can be in S, and x+1 and x+2 cannot. If we can cover the integers starting from 3 with such triples {x, x+1, x+2}, where each triple can contain at most one element of S (the x), then the density of S would be 1/3 beyond some point.But since 1 and 2 are already in S, they might interfere with this blocking. For example, 1 + x must not be in S, so if we have an element x, then 1 + x is excluded. Similarly, 2 + x is excluded. But if x is large enough, 1 + x and 2 + x might be in the next triple.Wait, let's formalize this. Suppose that starting from some number m, every number x ≥m is part of a triple {x, x+1, x+2}, and each triple can contain at most one element of S. Then the number of elements in S from m onwards would be at most the number of triples, which is approximately (k - m)/3. Adding the elements before m gives a constant, leading to the overall bound.But how do we handle the interaction between the elements? For example, if x is in S, then x +1 and x +2 are excluded. But x +1 might be part of another triple. Let me try to construct such a covering.Suppose we start at m=3. The triples would be {3,4,5}, {6,7,8}, {9,10,11}, etc. But 1 and 2 are already in S. Then:- 1 +3=4 is excluded.- 1 +4=5 is excluded.- 1 +5=6 is excluded.- 2 +3=5 is excluded.- 2 +4=6 is excluded.- 2 +5=7 is excluded.Wait, so if we consider the triple {3,4,5}, adding 3 to S would exclude 4 and 5. But 4 is excluded by 1 +3=4, and 5 is excluded by 2 +3=5. So actually, if we have 3 in S, then 4 and 5 are excluded. However, 4 is already excluded by 1 +3=4. Similarly, 5 is excluded by 2 +3=5. Therefore, if we include 3 in S, we exclude 4 and 5. However, in our original problem, we can't have 3 in S because 1 +2=3. Therefore, 3 is excluded. So the triple {3,4,5} can't have 3, and 4 is excluded because 1 +3=4, but 3 isn't in S. Wait, hold on. If 3 is not in S, is 4 excluded?No. The exclusion happens only if the sum is in S. Since 3 is not in S, 1 +3=4 doesn't exclude 4. But 4 can be excluded if there are two elements in S that sum to 4. For example, 1 +3=4, but since 3 is not in S, 4 is allowed. Similarly, 2 +2=4, but elements must be distinct, so 2 +2 is not considered. Therefore, 4 is allowed. However, in our earlier construction, we included 4 in S. Then, once 4 is in S, 1 +4=5 and 2 +4=6 are excluded. So 5 and 6 cannot be in S. Then, the next triple {6,7,8} is affected by previous exclusions. But since 6 is excluded, we can't have 6 in S, so we go to 7. Adding 7 to S excludes 8 and 9. Then {9,10,11}: 9 is excluded, so add 10, which excludes 11 and 12, etc.So, in this case, starting from 4, each element x we add excludes x +1 and x +2, effectively forcing us to take the next available number x +3. This creates the arithmetic progression with difference 3. But how does this lead to the bound?If we model the sequence as starting with 1,2,4,7,10,... where after 4, each term is 3 more than the previous, then the number of terms less than k is:- 1,2 (count=2) for k ≥2.- Plus the number of terms from 4 up to k-1, stepping by 3. The number of these terms is floor((k -1 -4)/3) +1. Which is approximately (k -5)/3 +1 = (k -2)/3. So total approximate count is 2 + (k -2)/3 = k/3 + 4/3. But the problem allows k/3 +2, which is a bit more lenient.But in reality, the exact count would depend on k. For example, if k=7, the elements less than 7 are 1,2,4, which is 3 elements. The formula gives 7/3 +2 ≈4.333, which is more than 3. So it's okay. For k=10, elements less than 10 are 1,2,4,7: 4 elements. Formula gives 10/3 +2 ≈5.333. Still okay.But the key idea is that after the first two elements, every new element effectively reserves the next two numbers as excluded, leading to a density of 1/3. Therefore, the total number of elements less than k is about 2 + (k -2)/3, which is k/3 + 4/3. But the problem statement allows k/3 +2, which is a slightly higher bound. Therefore, our constructed sequence is actually within a tighter bound, but the problem allows a bit more, perhaps accounting for different constructions or initial terms.However, we need to prove that no matter how the sequence is constructed (following the given rules), the number of elements less than k can't exceed k/3 +2.Perhaps another approach is to consider that each element after the first two blocks two subsequent numbers. Therefore, the maximum number of elements up to k is 2 + (k -2 +2)/3 = k/3 + 2. Wait, let's think.If every element after the first two blocks two numbers ahead of it, then for each element x ≥4, we can associate it with two numbers x+1 and x+2 that are excluded. So, in the range from 4 to k-1, each element occupies a position and blocks the next two. Therefore, the number of elements in this range is at most ceil((k -4)/3). Adding the initial two elements and 4, but maybe this is getting messy.Alternatively, using the concept of induction. Suppose that for all natural numbers up to k, the bound holds. Then for k +1, we can show that adding a new element doesn't violate the bound. But induction might be tricky here because the choice of elements affects future counts.Alternatively, let's use the following argument. For the sequence S, define that for any x in S, x ≥3, the numbers x +1 and x +2 are not in S. Because if x is in S, then 1 +x and 2 +x must not be in S. Therefore, for each x in S with x ≥3, we can associate two numbers x +1 and x +2 that are excluded from S.Additionally, since 1 and 2 are in S, the numbers 1 +1=2 (but 1 and 1 are the same element, so not considered), 1 +2=3 (which is excluded), and 2 +2=4 (again, same element). So, the presence of 1 and 2 excludes 3 (due to 1 +2=3). Therefore, 3 cannot be in S.Now, let's consider the set S' = S {1,2}. Then, S' is a subset of the positive integers starting from 3, with the property that for any x in S', x +1 and x +2 are not in S. Moreover, S' cannot contain 3 (as explained). Therefore, S' is a subset where no two elements are consecutive or have a difference of 2. Wait, no. If x and y are elements of S', then x +1 and x +2 are excluded, but y could be x +3, for example.But more precisely, if x is in S', then x +1 and x +2 cannot be in S'. Therefore, S' is a set where no two elements are within 2 of each other. That is, S' is a subset with the property that any two elements are at least 3 apart. Therefore, the maximum density of such a set is 1/3, since you can take every third number.Therefore, the size of S' up to k is at most floor((k -3)/3) +1. Which is approximately k/3 -1 +1 =k/3. Therefore, the total size of S up to k is 2 (for 1 and 2) plus at most k/3. Hence, total is at most k/3 +2. Which is exactly the bound we need.Wait, this seems promising. Let me check again.If we remove 1 and 2 from S, the remaining elements form a set S' where:- 3 is excluded (since 1+2=3).- For any x in S', x +1 and x +2 are excluded from S.Therefore, S' is a subset of integers ≥4 with no two elements within 2 of each other. In other words, elements of S' are at least 3 apart.Therefore, the maximum number of elements in S' less than k is at most floor((k -4)/3) +1. For example, if k=10, then (10 -4)/3=2, so floor(2)+1=3 elements: 4,7,10. But 10 is not less than 10, so actually up to 9: 4,7. Which is 2 elements. Hmm, floor((9 -4)/3) +1= floor(5/3)+1=1+1=2, correct.So in general, for k ≥4, the number of elements in S' less than k is at most floor((k -4)/3) +1. Then the total number of elements in S less than k is 2 (for 1 and 2) plus floor((k -4)/3) +1.Let's compute this:floor((k -4)/3) +1 = floor((k -4 +3)/3) = floor((k -1)/3).Wait, no. Let's do it step by step.Number of elements in S' less than k: starting from 4 up to k-1, with each element at least 3 apart. The number of such elements is the largest integer m such that 4 +3(m -1) <k.So 4 +3(m -1) <k => 3(m -1) <k -4 => m -1 < (k -4)/3 => m < (k -4)/3 +1. Therefore, m ≤ floor((k -4)/3) +1.Therefore, the number of elements in S' less than k is ≤ floor((k -4)/3) +1.Adding the two initial elements, total elements in S less than k is ≤2 + floor((k -4)/3) +1= floor((k -4)/3) +3.Now, let's compare this to the desired bound of k/3 +2.floor((k -4)/3) +3 ≤ (k -4)/3 +3= k/3 -4/3 +3= k/3 +5/3.But 5/3 is approximately 1.666, which is less than 2. So floor((k -4)/3) +3 ≤ k/3 +5/3 <k/3 +2 for k≥1. But wait, this seems contradictory, because floor((k -4)/3) +3 can be equal to floor(k/3 -4/3) +3= floor(k/3) -1 +3= floor(k/3) +2. For example, if k=7:floor((7 -4)/3) +3= floor(3/3)+3=1 +3=4. But k/3 +2≈7/3 +2≈4.333, so floor(k/3) +2=2 +2=4. Which is equal. For k=8:floor((8 -4)/3) +3= floor(4/3) +3=1 +3=4. k/3 +2≈8/3 +2≈4.666. So floor(k/3)+2=2 +2=4. So it seems that floor((k -4)/3) +3= floor(k/3) +2.Indeed, let's verify:floor((k -4)/3) +3= floor(k/3 -4/3) +3.If we write k=3m +r, where r=0,1,2.Then k/3 -4/3= m +r/3 -4/3.floor(k/3 -4/3)= m -2 if r=0 (since m -4/3 is not integer), wait:Wait, for k=3m +0:k/3 -4/3= m -4/3. floor(m -4/3)=m -2 if m ≥2, since m -4/3 is between m -1 and m -2.Wait, this seems confusing. Let's take specific examples.If k=6=3*2 +0:floor((6 -4)/3) +3= floor(2/3) +3=0 +3=3.floor(k/3) +2=2 +2=4. Wait, but 3≠4. Hmm, this contradicts. Wait, k=6:Elements in S less than 6: 1,2,4. So count=3. The formula floor((6 -4)/3) +3=0 +3=3. Correct. The bound k/3 +2=6/3 +2=4. So 3 ≤4, which holds. But floor(k/3) +2=2 +2=4. So the actual count is 3, which is less than 4.But our previous derivation gave that the number of elements is ≤ floor((k -4)/3) +3, which for k=6 is 3, and the bound is 4. So it's within the bound.Similarly, for k=7:floor((7 -4)/3)=1, so 1 +3=4. The actual elements less than7:1,2,4. Count=3 <4. So still within.For k=10:floor((10-4)/3)=2, so 2 +3=5. Actual elements less than10:1,2,4,7. Count=4 ≤5.For k=13:floor((13-4)/3)=3, so 3 +3=6. Actual elements less than13:1,2,4,7,10. Count=5 ≤6.So it seems that floor((k -4)/3) +3 is always ≤k/3 +2. Because floor((k -4)/3) ≤(k -4)/3, so floor((k -4)/3) +3 ≤(k -4)/3 +3= k/3 -4/3 +3= k/3 +5/3 <k/3 +2. Therefore, the total number of elements is ≤k/3 +5/3, which is less than k/3 +2. But the problem allows up to k/3 +2. So the bound is satisfied.Therefore, by considering that after removing the initial elements 1 and 2, the remaining elements form a set where each element is at least 3 apart, leading to a maximum of floor((k -4)/3) +1 elements, then adding back the two initial elements gives a total of floor((k -4)/3) +3 elements, which is ≤k/3 +5/3 <k/3 +2. However, the problem statement allows k/3 +2, which is a slightly higher upper bound, so our analysis shows that the actual number is even less than that. Therefore, the bound holds.Alternatively, maybe there's a different way to account for the elements, such that the total number is ≤k/3 +2. For example, considering that each element after the first two blocks three numbers: itself and the next two. But since the first two elements are 1 and 2, they block 3 (since 1+2=3). Then each subsequent element x blocks x+1 and x+2. Therefore, each element beyond the first two is responsible for three numbers: x, x+1, x+2, where x is included in the sequence, and x+1, x+2 are excluded. Therefore, the total number of elements up to k is at most 2 + floor(k/3). Because each trio of numbers {3m+1, 3m+2, 3m+3} can contain at most one element from the sequence. Wait, but this might not align correctly.Wait, if we partition the numbers starting from 3 into blocks of 3: {3,4,5}, {6,7,8}, {9,10,11}, etc. Each block can contain at most one element of the sequence. Because if you have an element in the block, say 4, then 4+1=5 and 4+2=6 are excluded. So in the block {3,4,5}, if 4 is in the sequence, then 5 is excluded. If 3 were in the sequence (but it's not allowed because 1+2=3), then 4 and 5 would be excluded. Therefore, each block can have at most one element. However, since the first block {3,4,5} cannot have 3, and 4 or 5 might be in the sequence. For example, in our constructed sequence, 4 is included, which excludes 5 and 6. Then the next block {6,7,8} can have 7, which excludes 8 and9. So each block {3m,3m+1,3m+2} can have at most one element (3m+1 in our constructed sequence). Therefore, the number of elements from 3 onwards is at most the number of blocks, which is floor(k/3). Therefore, total elements up to k is at most 2 + floor(k/3) +1 (if k is in a new block). Wait, this is getting confusing.Alternatively, using the idea that each element x ≥3 occupies a position and blocks the next two positions. Therefore, the number of elements in the sequence from 3 upwards is at most ceil((k -3)/3). Adding the two initial elements gives total ≤2 + ceil((k -3)/3). Let's check:For k=6: ceil((6-3)/3)=ceil(1)=1. Total=3. Actual elements less than6:1,2,4. Count=3. Okay.For k=7: ceil(4/3)=2. Total=4. Actual elements less than7:1,2,4. Count=3 ≤4. Okay.For k=10: ceil(7/3)=3. Total=5. Actual elements less than10:1,2,4,7. Count=4 ≤5. Okay.But ceil((k -3)/3) = ceil(k/3 -1). Which is equal to ceil(k/3) -1 if k/3 is not an integer, otherwise ceil(k/3) -1= k/3 -1. Therefore, 2 + ceil(k/3) -1=1 + ceil(k/3). But 1 + ceil(k/3) ≤k/3 +2, since ceil(k/3) ≤k/3 +1. Therefore, 1 + ceil(k/3) ≤k/3 +2. Hence, the total number of elements is ≤k/3 +2, as required.Therefore, combining these observations:- The elements 1 and 2 are in the sequence.- Every subsequent element x blocks x+1 and x+2 from being in the sequence.- Therefore, starting from 3, every third number can be in the sequence.- Hence, the number of elements from 3 upwards less than k is at most ceil((k -3)/3).- Adding the two initial elements gives a total of ceil((k -3)/3) +2.- Since ceil((k -3)/3) ≤k/3, we have ceil((k -3)/3) +2 ≤k/3 +2.Therefore, the bound holds.Thus, the number of elements less than k in the sequence is at most k/3 +2.</think>To prove that for any natural number ( k ), the number of elements in the sequence less than ( k ) is at most ( frac{k}{3} + 2 ), we analyze the structure and constraints of the sequence:1. Initial Elements: The sequence starts with ( 1 ) and ( 2 ). These elements exclude ( 3 ) (since ( 1 + 2 = 3 )).2. Blocking Rule: Each subsequent element ( x ) added to the sequence blocks ( x + 1 ) and ( x + 2 ) from being in the sequence (to prevent sums with ( 1 ) and ( 2 ), respectively).3. Residue Classes: After ( 1 ) and ( 2 ), elements must be spaced such that no two elements are within ( 2 ) units of each other. This leads to a pattern where elements are at least ( 3 ) apart starting from ( 4 ).4. Density Argument: Starting from ( 4 ), each element occupies a position and blocks the next two. This results in at most ( leftlfloor frac{k - 4}{3} rightrfloor + 1 ) elements from ( 4 ) to ( k-1 ).5. Total Count: Adding the initial elements ( 1 ) and ( 2 ), the total number of elements less than ( k ) is: [ 2 + leftlfloor frac{k - 4}{3} rightrfloor + 1 leq frac{k}{3} + 2 ] This holds because ( leftlfloor frac{k - 4}{3} rightrfloor leq frac{k - 4}{3} ), leading to: [ 2 + frac{k - 4}{3} + 1 = frac{k}{3} + frac{5}{3} < frac{k}{3} + 2 ]Thus, the number of elements less than ( k ) in the sequence is bounded by ( boxed{frac{k}{3} + 2} ).]

🔑:Solution. Since the first two elements are 1 and 2, 3 is not an element of the sequence, and out of the numbers 4 and 5, at most one can be an element of the sequence. If n geq 5 (a natural number) is in the sequence, then n-1 and n-2 cannot be, because (n-1)+1=n and (n-2)+2=n. That is, out of three consecutive numbers greater than 2, at most one can be in the sequence.Thus, if k is divisible by 3, we group the numbers from 4 to k as follows:4,5,6 ; quad 7,8,9 ; quad 10,11,12 ; quad ldots quad k-2 ; k-1 ; kIf k is not divisible by 3, then clearly the last group will contain 1 element (k) or two elements (k-1 ; k).For k divisible by 3, there are frac{k-3}{3} groups,if k gives a remainder of 1 when divided by 3, then there are frac{k-1}{3} groups,and for a remainder of 2, there are frac{k-2}{3} groups.Each of these is less than frac{k}{3}, so we can say that there are at most frac{k}{3} such groups, and thus the total number of elements less than k is at most 2+frac{k}{3}.

🧠:[<think>Okay, let me try to figure out this sequence problem. So, the problem says we have a sequence {a_n} where a₁ = -1, a₂ > a₁, and the absolute difference between consecutive terms is |a_{n+1} - a_n| = 2^n. Also, the subsequence of odd-indexed terms {a_{2n-1}} is monotonically decreasing, and the subsequence of even-indexed terms {a_{2n}} is monotonically increasing. We need to find the general formula for a_n.First, let me write down what's given:1. a₁ = -12. a₂ > a₁ (so a₂ > -1)3. |a_{n+1} - a_n| = 2^n for all natural numbers n4. The odd subsequence {a_{2n-1}} is decreasing5. The even subsequence {a_{2n}} is increasingHmm. Since the absolute difference between terms is 2^n, each term is either 2^n more or less than the previous term. But the direction (whether we add or subtract) depends on the conditions given about the subsequences.First, let's try to compute the first few terms to see a pattern.We know a₁ = -1.For n = 1: |a₂ - a₁| = 2¹ = 2. So a₂ = a₁ ± 2. But since a₂ > a₁, and a₁ is -1, so a₂ must be -1 + 2 = 1. Because if we subtract 2, we get -3, which is less than -1, which contradicts a₂ > a₁. So a₂ = 1.Now, n = 2: |a₃ - a₂| = 2² = 4. So a₃ = a₂ ± 4. Let's see. The subsequence of odd terms, which includes a₁, a₃, a₅, etc., must be decreasing. Since a₁ = -1 and a₃ is the next odd term, to be decreasing, a₃ must be less than a₁. But wait, a₁ is -1. So a₃ < -1. Since a₂ is 1, then a₃ can be 1 + 4 = 5 or 1 - 4 = -3. But since a₃ needs to be less than -1, a₃ must be -3.So a₃ = -3.Next, n = 3: |a₄ - a₃| = 2³ = 8. So a₄ = a₃ ± 8. The even subsequence {a₂, a₄, a₆, ...} must be increasing. Since a₂ = 1, a₄ must be greater than 1. Since a₃ = -3, then a₄ can be -3 + 8 = 5 or -3 - 8 = -11. But a₄ needs to be greater than 1, so a₄ = 5.So a₄ = 5.Now, n = 4: |a₅ - a₄| = 2⁴ = 16. So a₅ = a₄ ± 16. The odd subsequence must be decreasing, so a₅ < a₃ = -3. Since a₄ = 5, a₅ can be 5 + 16 = 21 or 5 - 16 = -11. But we need a₅ < -3, so a₅ = -11.So a₅ = -11.n = 5: |a₆ - a₅| = 2⁵ = 32. So a₆ = a₅ ± 32. The even subsequence {a₂, a₄, a₆, ...} is increasing, so a₆ must be greater than a₄ = 5. Since a₅ = -11, then a₆ can be -11 + 32 = 21 or -11 -32 = -43. To be greater than 5, a₆ must be 21.So a₆ = 21.Continuing, n = 6: |a₇ - a₆| = 2⁶ = 64. So a₇ = 21 ± 64. The odd subsequence is decreasing, so a₇ < a₅ = -11. Therefore, a₇ must be 21 - 64 = -43 (since 21 + 64 = 85 is greater than -11). So a₇ = -43.Similarly, n = 7: |a₈ - a₇| = 2⁷ = 128. a₈ = -43 ± 128. The even subsequence must be increasing, so a₈ > a₆ =21. Therefore, a₈ must be -43 + 128 = 85 (since -43 -128 = -171 < 21). So a₈ =85.So let me list out the terms we have so far:n : a_n1 : -12 : 13 : -34 : 55 : -116 :217 :-438 :85Hmm. Let me check if the odd terms are decreasing: -1, -3, -11, -43,... yes, each is less than the previous. Even terms: 1,5,21,85,... increasing, which they are.Looking at the terms, maybe there's a pattern here. Let's see the odd terms:a₁ = -1a₃ = -3 = -1 - 2a₅ = -11 = -3 -8a₇ = -43 = -11 -32Wait, the differences here: from -1 to -3 is -2, then -8, then -32... which are 2, 8, 32. 2 is 2^1, 8 is 2^3, 32 is 2^5. Each time subtracting 2^{2k-1} ?Similarly, the even terms:a₂ =1a₄=5=1 +4a₆=21=5 +16a₈=85=21 +64The differences here: 4,16,64, which are 4=2^2, 16=2^4, 64=2^6. So each even term adds 2^{2k} to the previous even term.So for the odd terms, starting from a₁ = -1, each next odd term is subtracting 2^{2k-1}, where k starts from 1. Similarly, even terms starting from a₂=1, each next even term is adding 2^{2k} where k starts from 1.Wait, maybe we can express a_{2n-1} and a_{2n} in terms of their own sequences.Let me consider the odd indices first. Let’s define b_n = a_{2n -1}. Then we have:b₁ = a₁ = -1b₂ = a₃ = -3b₃ = a₅ = -11b₄ = a₇ = -43We can see that each term is obtained by subtracting 2^{2n -1} from the previous term? Let's check:From b₁ to b₂: -1 to -3, difference is -2 = -2^1From b₂ to b₃: -3 to -11, difference is -8 = -2^3From b₃ to b₄: -11 to -43, difference is -32 = -2^5So, the differences are -2^{2n -1} where n starts at 1 for each step.Wait, actually, for the first difference (n=1), we subtract 2^1, for the next (n=2), subtract 2^3, then n=3 subtract 2^5. So in general, the difference between b_n and b_{n+1} is -2^{2n -1}. Therefore, we can write b_n as:b_n = b₁ + Σ_{k=1}^{n-1} (-2^{2k -1})Similarly, for the even terms, let’s define c_n = a_{2n}c₁ = a₂ =1c₂ = a₄ =5c₃ =a₆=21c₄=a₈=85The differences here are 4,16,64,... which are 2^2, 2^4, 2^6,...So each difference is 2^{2k} where k starts at 1.So the differences between c_n and c_{n+1} is +2^{2k} where k =1,2,...So similarly, c_n = c₁ + Σ_{k=1}^{n-1} 2^{2k}So let's compute these sums.Starting with the odd terms:b_n = -1 + Σ_{k=1}^{n-1} (-2^{2k -1}) = -1 - Σ_{k=1}^{n-1} 2^{2k -1}The sum Σ_{k=1}^{n-1} 2^{2k -1} is a geometric series with first term 2^1 =2, ratio 4, and number of terms n-1.The sum of a geometric series is a1*(r^m -1)/(r -1), where a1 is the first term, r is the ratio, m the number of terms.So sum = 2*(4^{n -1} -1)/(4 -1) = 2*(4^{n -1} -1)/3Therefore, b_n = -1 - [2*(4^{n -1} -1)/3] = -1 - (2*4^{n -1} -2)/3 = (-3 -2*4^{n -1} +2)/3 = (-1 -2*4^{n -1})/3Hmm, let's check with n=1: b₁ should be -1. Plugging n=1: (-1 -2*4^{0})/3 = (-1 -2)/3 = -3/3 = -1. Correct.n=2: b₂ = (-1 -2*4^{1})/3 = (-1 -8)/3 = -9/3 = -3. Correct.n=3: (-1 -2*4²)/3 = (-1 -32)/3 = -33/3 = -11. Correct.n=4: (-1 -2*4³)/3 = (-1 -128)/3 = -129/3 = -43. Correct. So the formula works.Similarly for the even terms:c_n =1 + Σ_{k=1}^{n-1}2^{2k}Sum Σ_{k=1}^{n-1}2^{2k} = Σ 4^k from k=1 to n-1.This is a geometric series with a1=4, ratio=4, number of terms n-1.Sum =4*(4^{n-1} -1)/(4 -1) =4*(4^{n -1} -1)/3Therefore, c_n=1 + [4*(4^{n -1} -1)/3] =1 + (4^{n} -4)/3 = (3 +4^n -4)/3 = (4^n -1)/3Check with n=1: (4^1 -1)/3=3/3=1. Correct.n=2: (4² -1)/3=15/3=5. Correct.n=3: (4³ -1)/3=63/3=21. Correct.n=4: (4⁴ -1)/3=255/3=85. Correct.So now, we have:For odd indices: a_{2n -1} = (-1 -2*4^{n -1})/3For even indices: a_{2n} = (4^n -1)/3Now, we need to combine these into a single formula for a_n. Let's see.Let’s write n as either 2k -1 or 2k. Let's express in terms of n.For the odd case: if m =2n -1, then n=(m +1)/2. So substituting into the formula for a_m:a_m = [ -1 -2*4^{( (m +1)/2 ) -1 } ] /3 = [ -1 -2*4^{(m -1)/2 -0 } ] /3Wait, let me re-express:Original formula for a_{2n -1} is (-1 -2*4^{n -1})/3.Let’s let m =2n -1. Then n = (m +1)/2.So substituting:a_m = (-1 -2*4^{( (m +1)/2 ) -1 })/3 = (-1 -2*4^{(m -1)/2 })/3Similarly, for the even case: a_{2n} = (4^n -1)/3. Let m=2n, then n=m/2. So:a_m = (4^{m/2} -1)/3 = (2^m -1)/3. Wait, 4^{m/2} is (2²)^{m/2} =2^{m}. So yes, (2^m -1)/3.Therefore, combining both cases:If m is odd: a_m = (-1 -2*4^{(m -1)/2}) /3If m is even: a_m = (2^m -1)/3But maybe we can write this in a single formula using powers of -2 or something. Alternatively, we can express it using (-2)^n.Wait, let me check for m=1:If m is odd: a₁ = (-1 -2*4^{(1-1)/2}) /3 = (-1 -2*4^0)/3 = (-1 -2)/3 = -1. Correct.For m=2, even: a₂=(2² -1)/3=(4 -1)/3=1. Correct.m=3, odd: (-1 -2*4^{(3-1)/2}) /3 = (-1 -2*4^1)/3=(-1 -8)/3=-3. Correct.m=4, even: (2^4 -1)/3=15/3=5. Correct.So, this seems to work. But how to combine this into a single formula?Notice that for odd m, we can write (-1 -2*4^{(m-1)/2})/3. Let's manipulate this:Note that 4^{(m-1)/2} =2^{m -1}So, (-1 -2*2^{m -1})/3 = (-1 -2^m)/3Wait, 2*2^{m -1}=2^{m}, so yes, (-1 -2^m)/3. But wait, when m is odd:For m=1: (-1 -2^1)/3=(-1 -2)/3=-1. Correct.m=3: (-1 -2^3)/3=(-1 -8)/3=-3. Correct.m=5: (-1 -2^5)/3=(-1 -32)/3=-11. Correct.Similarly, for even m:(2^m -1)/3So, if we can write a general formula that distinguishes between even and odd m, we can use:a_n = [ (-1 -2^n)/3 ] if n is odd,a_n = [ (2^n -1)/3 ] if n is even.But can we combine these into a single expression?Alternatively, using (-2)^n. Let's see:For odd n:(-2)^n = -2^nSo (-2)^n -1 = -2^n -1Then [ (-2)^n -1 ] /3 = (-2^n -1)/3 = same as the odd case.For even n:(-2)^n = 4^{n/2} =2^nSo [ (-2)^n -1 ] /3 = (2^n -1)/3 = same as the even case.Wait, that works!So, combining both cases:a_n = [ (-2)^n -1 ] /3Let’s check this:For n=1: [ (-2)^1 -1 ] /3 = (-2 -1)/3 = -1. Correct.n=2: [ (-2)^2 -1 ] /3 = (4 -1)/3 =1. Correct.n=3: [ (-2)^3 -1 ] /3 = (-8 -1)/3=-3. Correct.n=4: [ (-2)^4 -1 ] /3=(16 -1)/3=5. Correct.n=5: [ (-2)^5 -1]/3=(-32 -1)/3=-11. Correct.n=6: [ (-2)^6 -1]/3=(64 -1)/3=21. Correct.Yes! So this formula works for all n. Therefore, the general term is a_n = [ (-2)^n -1 ] /3.But let me verify once more with n=7 and n=8.n=7: [ (-2)^7 -1 ] /3=(-128 -1)/3=-129/3=-43. Correct.n=8: [ (-2)^8 -1 ] /3=(256 -1)/3=255/3=85. Correct. Perfect.Therefore, the answer is a_n = frac{(-2)^n -1}{3}Final AnswerThe general term formula for the sequence is boxed{dfrac{(-2)^n - 1}{3}}.</think>Given the sequence {a_n} satisfying a_1 = -1, a_2 > a_1, and |a_{n+1} - a_n| = 2^n (where n in mathbb{N}^*), with the subsequence {a_{2n-1}} being monotonically decreasing and the subsequence {a_{2n}} being monotonically increasing, we need to find the general term formula for the sequence {a_n}.1. Initial Terms Calculation: - a_1 = -1 - a_2 = 1 (since a_2 > a_1 and |a_2 - a_1| = 2) - a_3 = -3 (since {a_{2n-1}} is decreasing and |a_3 - a_2| = 4) - a_4 = 5 (since {a_{2n}} is increasing and |a_4 - a_3| = 8) - Continuing this pattern, we calculate the next terms: a_5 = -11, a_6 = 21, a_7 = -43, a_8 = 85, etc.2. Pattern Recognition: - The odd-indexed terms a_{2n-1} form a sequence: -1, -3, -11, -43, ldots - The even-indexed terms a_{2n} form a sequence: 1, 5, 21, 85, ldots3. General Term Derivation: - For odd indices a_{2n-1}: The terms can be expressed as b_n = -1 - sum_{k=1}^{n-1} 2^{2k-1}. This simplifies to b_n = frac{(-2)^{2n-1} - 1}{3}. - For even indices a_{2n}: The terms can be expressed as c_n = 1 + sum_{k=1}^{n-1} 2^{2k}. This simplifies to c_n = frac{2^{2n} - 1}{3}.4. Combining into a Single Formula: - Observing that both sequences can be combined using (-2)^n: - For odd n: a_n = frac{(-2)^n - 1}{3} - For even n: a_n = frac{(-2)^n - 1}{3}Thus, the general term formula for the sequence {a_n} is:[boxed{dfrac{(-2)^n - 1}{3}}],<think>Okay, let me try to figure out the general term formula for this sequence. Hmm, the problem says that we have a sequence {a_n} where a₁ = -1, and a₂ is greater than a₁. The absolute difference between consecutive terms is 2ⁿ, meaning |a_{n+1} - a_n| = 2ⁿ for each natural number n. Also, the subsequence of odd-indexed terms {a_{2n-1}} is monotonically decreasing, and the subsequence of even-indexed terms {a_{2n}} is monotonically increasing. We need to find a formula for a_n.First, let's note the starting point: a₁ = -1. Then a₂ has to be greater than a₁, and the difference between a₂ and a₁ is 2¹ = 2. So since a₂ > a₁, that means a₂ = a₁ + 2 = -1 + 2 = 1. Okay, so a₂ = 1.Next, let's move to a₃. The difference |a₃ - a₂| = 2² = 4. Now, we need to consider the subsequence of odd terms, which is {a₁, a₃, a₅, ...}. This subsequence must be monotonically decreasing. Since a₁ = -1, the next term in this subsequence is a₃. To be decreasing, a₃ must be less than a₁, so a₃ < -1. But the difference from a₂ is |a₃ - a₂| = 4. Since a₃ < -1 and a₂ = 1, the difference would be 1 - a₃ = 4, which implies a₃ = 1 - 4 = -3. Alternatively, if we thought a₃ - a₂ = -4, then a₃ = a₂ -4 = 1 - 4 = -3. Either way, a₃ = -3.Now, moving to a₄. The difference |a₄ - a₃| = 2³ = 8. The even-indexed subsequence {a₂, a₄, a₆, ...} is increasing. Since a₂ = 1, the next term a₄ must be greater than a₂. So a₄ > 1. The difference |a₄ - a₃| = 8, and since a₃ = -3, then a₄ = a₃ + 8 = -3 + 8 = 5. Alternatively, a₄ could be -3 -8 = -11, but that would make a₄ = -11, which is less than a₂ = 1, contradicting the requirement that the even subsequence is increasing. Hence, a₄ = 5.Continuing to a₅. The difference |a₅ - a₄| = 2⁴ = 16. The odd subsequence {a₁, a₃, a₅, ...} must be decreasing. The last term in the odd subsequence so far is a₃ = -3, so a₅ must be less than -3. The difference with a₄ = 5 is |a₅ - 5| = 16. To make a₅ < -3, we need a₅ = 5 - 16 = -11. Alternatively, if we take a₅ = 5 + 16 = 21, but that would not be less than -3, so definitely a₅ = -11.Similarly, a₆ is next. |a₆ - a₅| = 32. Since the even subsequence is increasing, a₆ must be greater than a₄ = 5. So a₆ = a₅ + 32 = -11 + 32 = 21. If we had subtracted, a₆ = a₅ - 32 = -43, which is less than 5, so that's invalid. So a₆ = 21.So far, the terms are: a₁=-1, a₂=1, a₃=-3, a₄=5, a₅=-11, a₆=21, a₇=-43, a₈=85,... Hmm, let me check a₇ and a₈. For a₇: |a₇ - a₆| = 64. The odd subsequence requires a₇ < a₅ = -11. So a₇ = a₆ - 64 = 21 - 64 = -43. Then a₈ is |a₈ - a₇| = 128. Even subsequence needs a₈ > a₆ =21. So a₈ = a₇ + 128 = -43 +128=85. Yep, that works.Looking at these terms, let's see if there's a pattern. Let's list them:n: 1 2 3 4 5 6 7 8 ...a_n: -1, 1, -3, 5, -11, 21, -43, 85...Looking at the odd terms: -1, -3, -11, -43... Each term seems to be previous odd term multiplied by 3 and then subtract 2? Wait, -1*3 -2 = -5, but next term is -3. Hmm, maybe not.Wait, let's check differences:From -1 to -3: difference -2From -3 to -11: difference -8From -11 to -43: difference -32From -43 to -171 (if continued): difference -128So the differences are -2, -8, -32, -128,... which is multiplied by 4 each time. Similarly, the even terms:1, 5, 21, 85...Differences: 4, 16, 64,... also multiplied by 4 each time.So, for the odd terms, starting at -1, each term is previous term minus 2*4^{k} where k starts at 0. Similarly, even terms starting at 1, each term is previous term plus 4^{k+1} or something. Let's see.Wait, the differences for odd terms are -2, -8, -32, which is -2*1, -2*4, -2*16, so each time multiplied by 4. Similarly, even terms differences are 4, 16, 64, which is 4*1, 4*4, 4*16, etc. So each even term adds 4*4^{k}.So perhaps we can model the odd and even terms as separate geometric sequences.For the odd terms (indices 2k-1):a_1 = -1a_3 = -1 -2 = -3a_5 = -3 -8 = -11a_7 = -11 -32 = -43So each time, subtracting 2*4^{k-1}, where k is the term index in the odd subsequence.Similarly, for the even terms (indices 2k):a_2 = 1a_4 = 1 +4 =5a_6 =5 +16=21a_8=21 +64=85Each time adding 4^{k}, where k starts at 1.So for the odd subsequence: The recursive formula would be a_{2k+1} = a_{2k-1} - 2*4^{k-1}Similarly, for the even subsequence: a_{2k} = a_{2k-2} +4^{k-1}Alternatively, for the odd terms, starting at k=1, term 1: a₁=-1; term 2 (k=2): a₃=-3; term 3 (k=3): a₅=-11; etc. So each term is -1 -2*(4^{0} +4^{1} + ... +4^{k-2})Wait, because from a₁ to a₃: subtract 2*4⁰From a₃ to a₅: subtract 2*4¹From a₅ to a₇: subtract 2*4²So in general, a_{2k-1} = -1 -2*(4⁰ +4¹ + ... +4^{k-2})Similarly, the sum inside is a geometric series. The sum of 4⁰ +4¹ + ... +4^{n} is (4^{n+1} -1)/3.So if we have k terms, starting from 0 to k-2, the sum is (4^{k-1} -1)/3.So a_{2k-1} = -1 -2*( (4^{k-1} -1)/3 )Similarly, let's check for k=1: a₁ = -1 -2*( (4^{0} -1)/3 ) = -1 -2*(0/3)= -1, correct.For k=2: a₃ = -1 -2*( (4^{1} -1)/3 )= -1 -2*(3/3)= -1 -2= -3, correct.For k=3: a₅ = -1 -2*( (4² -1)/3 )= -1 -2*(15/3)= -1 -10= -11, correct. Good.Similarly, for even terms:a_2=1, a_4=5, a_6=21, a_8=85,...Each term is previous term plus 4^{k-1}, starting from a_2=1.So for a_{2k}, starting with k=1:a_{2} =1a_{4}=1 +4¹=5a_{6}=5 +4²=21a_{8}=21 +4³=85So in general, a_{2k} =1 +4¹ +4² +...+4^{k-1}Again, this is a geometric series sum. The sum is (4^{k} -1)/3.Therefore, a_{2k} = (4^{k} -1)/3.Check for k=1: (4 -1)/3=1, correct. For k=2: (16-1)/3=5, correct. For k=3: (64-1)/3=21, correct. So that works.Therefore, the odd terms can be written as a_{2k-1} = -1 -2*( (4^{k-1} -1)/3 ) = -1 - (2/3)(4^{k-1} -1 ) = -1 - (2/3)4^{k-1} + 2/3 = (-1 + 2/3) - (2/3)4^{k-1} = (-1/3) - (2/3)4^{k-1}Alternatively, factor out the -1/3:a_{2k-1} = - (1 + 2*4^{k-1} ) / 3Wait, let's compute:Starting with:a_{2k-1} = -1 - (2/3)(4^{k-1} -1 )= -1 - (2/3)4^{k-1} + 2/3= (-1 + 2/3) - (2/3)4^{k-1}= (-1/3) - (2/3)4^{k-1}= - (1 + 2*4^{k-1}) /3Alternatively, 4^{k-1} = (4^{k})/4, so:= - (1 + (4^{k}/2) ) /3Wait, maybe not. Let's see:But 4^{k-1} = (1/4)4^{k}, so:= - (1 + 2*(1/4)4^{k} ) /3= - (1 + (4^{k}/2) ) /3Hmm, not sure if that helps.Alternatively, let's express in terms of exponents:Note that 4^{k-1} = (2^2)^{k-1} = 2^{2(k-1)}.So:a_{2k-1} = - (1 + 2*2^{2(k-1)} ) /3= - (1 + 2^{2k -1} ) /3But perhaps not. Let's check for k=1:a₁ = - (1 + 2^{1} ) /3 = -3/3 = -1, correct.k=2: a₃= - (1 + 2^{3} ) /3 = -9/3= -3, correct.k=3: a₅= - (1 + 2^{5} ) /3= -33/3= -11, correct. So yes, that works. So a_{2k-1} = - (1 + 2^{2k -1} ) /3.Wait, let's verify:For k=1: 2k-1=1, so formula: - (1 + 2^{1}) /3 = -3/3=-1. Correct.k=2: 2k-1=3: - (1 + 2^3)/3= -9/3=-3. Correct.k=3: 2k-1=5: - (1 +2^5)/3= -33/3=-11. Correct. So indeed, a_{2k-1}= - (1 +2^{2k-1})/3.Similarly, the even terms a_{2k}= (4^k -1)/3.Now, we need to write a general formula for a_n, where n can be either even or odd. Let's consider n as either 2k-1 or 2k. Let's express k in terms of n.If n is odd: n=2k-1 ⇒ k=(n+1)/2If n is even: n=2k ⇒ k= n/2So for n odd: a_n= a_{2k-1}= - (1 +2^{2k-1})/3. Substitute k=(n+1)/2:So 2k-1= 2*( (n+1)/2 ) -1 = n+1 -1 =n. So a_n= - (1 +2^{n})/3.Wait, hold on, that seems too simple. Let's check:For n=1 (odd): a₁= - (1 +2^1)/3= -3/3=-1. Correct.n=3: a₃= - (1 +8)/3= -9/3=-3. Correct.n=5: a₅= - (1 +32)/3= -33/3=-11. Correct. So seems like for odd n, a_n= - (1 +2^n)/3.For even n: a_{2k}= (4^k -1)/3. Now, n=2k, so k=n/2. Therefore, 4^k=4^{n/2}=2^{n}. So:a_n= (2^{n} -1)/3.Check:n=2: (4 -1)/3=1. Correct.n=4: (16 -1)/3=5. Correct.n=6: (64 -1)/3=21. Correct. So yes, for even n, a_n= (2^n -1)/3.Therefore, combining both cases, we can write:a_n= { - (1 +2^n)/3, if n is odd; (2^n -1)/3, if n is even.}But the problem asks for a general term formula for the sequence {a_n}, so we need to write this as a single expression. To do that, we can use (-1)^n to handle the alternation between odd and even terms.Alternatively, note that for odd n, (-1)^n = -1, and for even n, (-1)^n =1. Let's see:If we write a_n= [ (2^n -1)/3 ] * (-1)^n ?Wait, let's test:For n=1 (odd): (-1)^1= -1, so [ (2 -1)/3 ]*(-1)= (1/3)*(-1)= -1/3. But a₁ is -1. Not matching.Hmm, not quite. Alternatively, perhaps:Looking at the odd terms: - (1 +2^n)/3. Let's factor out the negative sign:= (-1 -2^n)/3.Even terms: (2^n -1)/3.So the difference between odd and even terms is that in the numerator, the sign of 1 and 2^n are opposite. So if we can write a formula that alternates the sign of 1 and 2^n based on n's parity.Wait, perhaps we can write:a_n= [ (-1)^n *1 +2^n ] /3 * (-1)^n ?Wait, let me see:For odd n, let's say n=2k-1. Then (-1)^n= -1. So:[ (-1)^{2k-1} *1 +2^{2k-1} ] /3 * (-1)^{2k-1} = [ (-1)*1 +2^{2k-1} ] /3 * (-1) = [ -1 +2^{2k-1} ] /3 * (-1) = [2^{2k-1} -1]/3 * (-1) = (-2^{2k-1} +1)/3= (1 -2^{2k-1}) /3= - (2^{2k-1} -1)/3. But the odd term is supposed to be - (1 +2^n)/3. Wait, that doesn't match. Hmm, perhaps not.Alternatively, let's think of combining the two cases:For odd n: a_n= - (1 +2^n)/3.For even n: a_n= (2^n -1)/3.Note that 2^n -1 and -(1 +2^n) can be written as (-1)^n*(2^n + (-1)^n). Let's check:For even n, (-1)^n=1: 2^n +1, multiplied by 1, so (2^n +1)/3. But even terms are (2^n -1)/3. Not quite.Wait, maybe:Let me compute a_n for even and odd:Odd n: a_n= - (1 +2^n)/3= (-1 -2^n)/3= [ -2^n -1 ] /3.Even n: a_n= (2^n -1)/3= [2^n -1]/3.So if we can write a_n= [ (-1)^n *2^n -1 ] /3.Check for even n: (-1)^n=1, so (2^n -1)/3. Correct.For odd n: (-1)^n=-1, so (-2^n -1)/3. Which is the same as - (2^n +1)/3. Correct.Yes! So that works. Therefore, the general term is a_n= [ (-1)^n *2^n -1 ] /3.Let me verify with the known terms:n=1: [ (-1)^1*2^1 -1 ] /3= [ -2 -1 ]/3= -3/3=-1. Correct.n=2: [1*4 -1]/3=3/3=1. Correct.n=3: [ -1*8 -1 ]/3= (-9)/3=-3. Correct.n=4: [1*16 -1]/3=15/3=5. Correct.n=5: [ -1*32 -1 ]/3= (-33)/3=-11. Correct.n=6: [1*64 -1]/3=63/3=21. Correct.Perfect. So the formula a_n= [ (-1)^n *2^n -1 ] /3 gives the correct terms. So the answer is boxed{a_n = dfrac{(-1)^n cdot 2^n -1}{3}}.Final AnswerThe general term formula for the sequence is boxed{a_n = dfrac{(-1)^n cdot 2^n -1}{3}}.</think>Given the sequence {a_n} with a_1 = -1, a_2 > a_1, and |a_{n+1} - a_n| = 2^n for n in mathbb{N}^*, where the subsequence {a_{2n-1}} is monotonically decreasing and the subsequence {a_{2n}} is monotonically increasing, we need to find the general term formula for {a_n}.1. Initial Terms Calculation: - a_1 = -1 - a_2 = a_1 + 2 = 1 - a_3 = a_2 - 4 = -3 - a_4 = a_3 + 8 = 5 - a_5 = a_4 - 16 = -11 - a_6 = a_5 + 32 = 21 - a_7 = a_6 - 64 = -43 - a_8 = a_7 + 128 = 852. Pattern Recognition: - The odd-indexed terms: -1, -3, -11, -43, ldots form a geometric sequence with differences multiplied by 4 each time. - The even-indexed terms: 1, 5, 21, 85, ldots form a geometric sequence with differences multiplied by 4 each time.3. General Term Derivation: - For odd n = 2k - 1: [ a_{2k-1} = -frac{1 + 2^{2k-1}}{3} ] - For even n = 2k: [ a_{2k} = frac{2^{2k} - 1}{3} ]4. Combining Both Cases: - Using (-1)^n to handle the alternation between odd and even terms: [ a_n = frac{(-1)^n cdot 2^n - 1}{3} ]Verification with initial terms confirms the correctness of the formula. Therefore, the general term formula for the sequence is:[boxed{a_n = dfrac{(-1)^n cdot 2^n -1}{3}}]]

🔑:There are two methods to solve this problem:Method 1: EnumerationBy listing out the first few terms, we have:- a_1 = -1- a_2 = 1 (since a_2 > a_1)- a_3 = a_2 - 2^2 = 1 - 4 = -3- a_4 = a_3 + 2^3 = -3 + 8 = 5- a_5 = a_4 - 2^4 = 5 - 16 = -11- a_6 = a_5 + 2^5 = -11 + 32 = 21- And so on...From these values, we can observe a pattern:a_{n+1} - a_n = (-1)^{n+1}2^n.Therefore, we can express a_n as the sum of the differences of consecutive terms:begin{align*}a_n &= (a_n - a_{n-1}) + (a_{n-1} - a_{n-2}) + ldots + (a_2 - a_1) + a_1 &= (-1)^n cdot 2^{n-1} + (-1)^{n-1} cdot 2^{n-2} + ldots - 2^2 + 2 -1 &= frac{(-1)((-2)^n - 1)}{3} &= frac{(-2)^n - 1}{3}.end{align*}This expression is derived by summing up the geometric series, and the final general term formula for the sequence is:a_n = boxed{frac{(-2)^n - 1}{3}}.Method 2: Analyzing the Sign of DifferencesSince {a_{2n-1}} is decreasing, we have:a_{2n+1} - a_{2n} = -2^{2n},because their difference must be negative. Similarly, since {a_{2n}} is increasing:a_{2n} - a_{2n-1} = 2^{2n-1},because their difference must be positive. Therefore:a_{n+1} - a_n = (-1)^{n+1} 2^n.Following the same steps as in Method 1, we arrive at the same conclusion:a_n = boxed{frac{(-2)^n - 1}{3}}.