💡Alright, so I have this inequality involving natural logarithms and some quadratic terms. It says:4 ln x + 2 ln(2y) ≥ x² + 8y - 4And I need to figure out which of the options A, B, C, D are correct based on this inequality. Let me try to break it down step by step.First, I remember that logarithms can be combined using their properties. Specifically, ln(a) + ln(b) = ln(ab), and coefficients in front of the logarithm can be turned into exponents. So, let me rewrite the left side of the inequality.4 ln x can be written as ln(x⁴), and 2 ln(2y) can be written as ln((2y)²) which is ln(4y²). So, combining these, the left side becomes ln(x⁴ * 4y²). Let me write that:ln(4x⁴y²) ≥ x² + 8y - 4Hmm, okay. So, the inequality is now:ln(4x⁴y²) ≥ x² + 8y - 4I wonder if I can simplify this further or maybe make a substitution to make it easier. Let me think. Maybe I can let u = x² and v = 2y. Then, 4x⁴y² would be (x²)² * (2y)², which is u²v². So, ln(u²v²) = 2 ln(uv). Let me try that substitution.Let u = x² and v = 2y. Then, the inequality becomes:2 ln(uv) ≥ u + 4v - 4Dividing both sides by 2 to simplify:ln(uv) ≥ (u + 4v - 4)/2Which is:ln(uv) ≥ (u/2) + 2v - 2Hmm, not sure if that helps directly, but maybe. Alternatively, I recall that sometimes inequalities involving logarithms can be approached by considering the function f(t) = ln(t) - t + 1, which has a maximum at t=1 and is concave. Maybe that can be useful here.Let me consider the function f(t) = ln(t) - t + 1. Its derivative is f’(t) = 1/t - 1, which is positive when t < 1 and negative when t > 1. So, f(t) has a maximum at t=1, where f(1) = 0. This function is concave, so it's always below its maximum.Looking back at the inequality, maybe I can express it in terms of f(t). Let me try to manipulate the inequality:ln(4x⁴y²) - (x² + 8y - 4) ≥ 0Let me split the logarithm:ln(4) + ln(x⁴) + ln(y²) - x² - 8y + 4 ≥ 0Simplify each term:ln(4) is a constant, ln(x⁴) is 4 ln x, ln(y²) is 2 ln y. So, substituting back:ln(4) + 4 ln x + 2 ln y - x² - 8y + 4 ≥ 0Wait, but the original inequality was 4 ln x + 2 ln(2y) ≥ x² + 8y - 4. Let me check if this aligns.Yes, because 2 ln(2y) is 2 ln 2 + 2 ln y, so 4 ln x + 2 ln(2y) is 4 ln x + 2 ln 2 + 2 ln y. So, the left side is 4 ln x + 2 ln 2 + 2 ln y, and the right side is x² + 8y - 4.So, bringing everything to the left:4 ln x + 2 ln 2 + 2 ln y - x² - 8y + 4 ≥ 0Which is the same as:ln(4) + 4 ln x + 2 ln y - x² - 8y + 4 ≥ 0So, that's consistent.Now, I think maybe I can group terms in a way that allows me to use the function f(t) = ln(t) - t + 1. Let me see.Let me consider terms involving x and terms involving y separately.First, for x:4 ln x - x²Similarly, for y:2 ln y - 8yAnd then the constants:ln(4) + 4So, the inequality becomes:(4 ln x - x²) + (2 ln y - 8y) + (ln 4 + 4) ≥ 0Hmm, maybe I can factor out some constants to make it look like f(t).For the x terms: 4 ln x - x² = 4(ln x - (x²)/4)Similarly, for y terms: 2 ln y - 8y = 2(ln y - 4y)So, substituting back:4(ln x - (x²)/4) + 2(ln y - 4y) + (ln 4 + 4) ≥ 0Let me denote:A = ln x - (x²)/4B = ln y - 4yThen, the inequality becomes:4A + 2B + (ln 4 + 4) ≥ 0Now, I can think of A and B as similar to the function f(t) = ln t - t + 1, but scaled differently.Wait, for A, it's ln x - (x²)/4. If I let u = x²/4, then ln x = (1/2) ln(4u). Hmm, not sure if that helps.Alternatively, maybe I can complete the square or find the maximum of each expression.Let me consider A = ln x - (x²)/4. Let's find its maximum.Take derivative of A with respect to x:dA/dx = (1/x) - (x/2)Set derivative equal to zero:(1/x) - (x/2) = 0Multiply both sides by 2x:2 - x² = 0 => x² = 2 => x = sqrt(2)So, the maximum of A occurs at x = sqrt(2). Let me compute A at this point:A = ln(sqrt(2)) - ( (sqrt(2))² ) /4 = (1/2) ln 2 - (2)/4 = (1/2) ln 2 - 1/2Similarly, for B = ln y - 4y.Take derivative of B with respect to y:dB/dy = (1/y) - 4Set derivative equal to zero:(1/y) - 4 = 0 => y = 1/4So, the maximum of B occurs at y = 1/4. Compute B at this point:B = ln(1/4) - 4*(1/4) = ln(1/4) - 1 = -ln 4 - 1So, now, going back to the inequality:4A + 2B + (ln 4 + 4) ≥ 0At the maximum points, A = (1/2) ln 2 - 1/2 and B = -ln 4 - 1.Plugging these in:4*(1/2 ln 2 - 1/2) + 2*(-ln 4 - 1) + (ln 4 + 4)Simplify each term:4*(1/2 ln 2) = 2 ln 24*(-1/2) = -22*(-ln 4) = -2 ln 42*(-1) = -2So, combining all terms:2 ln 2 - 2 - 2 ln 4 - 2 + ln 4 + 4Simplify:2 ln 2 - 2 - 2 ln 4 - 2 + ln 4 + 4Combine like terms:2 ln 2 - 2 ln 4 + ln 4 - 2 - 2 + 4Note that ln 4 = 2 ln 2, so:2 ln 2 - 2*(2 ln 2) + 2 ln 2 - 2 - 2 + 4Simplify:2 ln 2 - 4 ln 2 + 2 ln 2 - 4 + 4Combine the ln terms:(2 - 4 + 2) ln 2 = 0 ln 2 = 0Combine constants:-4 + 4 = 0So, total is 0.Therefore, at the maximum points x = sqrt(2), y = 1/4, the inequality becomes equality.So, this suggests that the inequality holds with equality when x = sqrt(2) and y = 1/4, and for other values, it might not hold or hold in a different way.Therefore, the only solution that satisfies the inequality is x = sqrt(2) and y = 1/4.Now, let's check the options:A: xy = sqrt(2)/4Compute xy: sqrt(2) * (1/4) = sqrt(2)/4. So, A is correct.B: x + y = sqrt(2)Compute x + y: sqrt(2) + 1/4 ≈ 1.414 + 0.25 ≈ 1.664, which is not equal to sqrt(2) ≈ 1.414. So, B is incorrect.C: x + 2y = 1/2 + sqrt(2)Compute x + 2y: sqrt(2) + 2*(1/4) = sqrt(2) + 1/2, which is equal to 1/2 + sqrt(2). So, C is correct.D: x² y = 1Compute x² y: (sqrt(2))² * (1/4) = 2 * (1/4) = 1/2, which is not equal to 1. So, D is incorrect.Therefore, the correct options are A and C.

💡First, I observe that the given expression is (sin 13^circ cos 17^circ + cos 13^circ sin 17^circ).I recall the trigonometric identity for the sine of a sum of two angles:[sin(A + B) = sin A cos B + cos A sin B]By comparing the given expression with the identity, I can set (A = 13^circ) and (B = 17^circ).Applying the identity:[sin(13^circ + 17^circ) = sin 30^circ]I know that (sin 30^circ = frac{1}{2}).Therefore, the value of the given expression is (frac{1}{2}).

💡Okay, so I have this problem where a sequence is defined recursively. It starts with x₀ = 7, and each subsequent term is given by the formula:x_{n+1} = (x_n² + 6x_n + 5)/(x_n + 7)I need to find the least positive integer m such that x_m ≤ 4 + 1/(2²⁵). Then, I have to determine which interval m falls into from the given options.First, I want to understand how this sequence behaves. It starts at 7, and each term is defined based on the previous one. Maybe I can compute the first few terms to see if there's a pattern or if it converges to a certain value.Let me compute x₁:x₁ = (7² + 6*7 + 5)/(7 + 7) = (49 + 42 + 5)/14 = 96/14 ≈ 6.857Okay, so x₁ is approximately 6.857, which is less than 7. Let's compute x₂:x₂ = (6.857² + 6*6.857 + 5)/(6.857 + 7)First, compute the numerator:6.857² ≈ 47.016*6.857 ≈ 41.14Adding them up with 5: 47.01 + 41.14 + 5 ≈ 93.15Denominator: 6.857 + 7 ≈ 13.857So x₂ ≈ 93.15 / 13.857 ≈ 6.727Hmm, it's decreasing again. Let me do one more term:x₃ = (6.727² + 6*6.727 + 5)/(6.727 + 7)Compute numerator:6.727² ≈ 45.256*6.727 ≈ 40.36Adding 5: 45.25 + 40.36 + 5 ≈ 90.61Denominator: 6.727 + 7 ≈ 13.727So x₃ ≈ 90.61 / 13.727 ≈ 6.596It's still decreasing. It seems like the sequence is approaching some limit. Maybe I can find the fixed point of this recursion. A fixed point would be a value x such that x = (x² + 6x + 5)/(x + 7).Let me solve for x:x(x + 7) = x² + 6x + 5x² + 7x = x² + 6x + 5Subtract x² from both sides:7x = 6x + 5Subtract 6x:x = 5Wait, so the fixed point is 5. But our sequence is starting at 7 and decreasing towards 5. However, the problem wants x_m ≤ 4 + 1/(2²⁵), which is approximately 4.000000000000000000000001. That's much less than 5. So, if the fixed point is 5, but we need to go below 4, that suggests that maybe the sequence doesn't just converge to 5 but perhaps oscillates or does something else?Wait, maybe I made a mistake. Let me check the fixed point again.Starting with x = (x² + 6x + 5)/(x + 7)Multiply both sides by (x + 7):x(x + 7) = x² + 6x + 5x² + 7x = x² + 6x + 5Subtract x²:7x = 6x + 5x = 5So, yes, the fixed point is indeed 5. So, if the sequence converges, it should approach 5. But the problem is asking for when it gets below 4 + something very small. That suggests that maybe the sequence doesn't just approach 5 but perhaps overshoots or something else is happening.Wait, maybe I need to analyze the behavior more carefully. Let me compute a few more terms to see.x₃ ≈ 6.596x₄ = (6.596² + 6*6.596 + 5)/(6.596 + 7)Compute numerator:6.596² ≈ 43.516*6.596 ≈ 39.58Adding 5: 43.51 + 39.58 + 5 ≈ 88.09Denominator: 6.596 + 7 ≈ 13.596x₄ ≈ 88.09 / 13.596 ≈ 6.476Still decreasing. Let me go further.x₅ ≈ (6.476² + 6*6.476 + 5)/(6.476 + 7)Compute numerator:6.476² ≈ 41.946*6.476 ≈ 38.86Adding 5: 41.94 + 38.86 + 5 ≈ 85.8Denominator: 6.476 + 7 ≈ 13.476x₅ ≈ 85.8 / 13.476 ≈ 6.365Hmm, it's decreasing, but the rate of decrease is slowing down. Let me try to see if I can model this recursion.Let me denote y_n = x_n - 5. So, shifting the sequence by 5. Then, x_n = y_n + 5.Substitute into the recursion:x_{n+1} = ( (y_n + 5)^2 + 6(y_n + 5) + 5 ) / (y_n + 5 + 7 )Simplify numerator:(y_n² + 10y_n + 25) + (6y_n + 30) + 5 = y_n² + 16y_n + 60Denominator: y_n + 12So,x_{n+1} = (y_n² + 16y_n + 60)/(y_n + 12)But x_{n+1} = y_{n+1} + 5, so:y_{n+1} + 5 = (y_n² + 16y_n + 60)/(y_n + 12)Multiply both sides by (y_n + 12):(y_{n+1} + 5)(y_n + 12) = y_n² + 16y_n + 60Expand the left side:y_{n+1}y_n + 12y_{n+1} + 5y_n + 60 = y_n² + 16y_n + 60Subtract 60 from both sides:y_{n+1}y_n + 12y_{n+1} + 5y_n = y_n² + 16y_nBring all terms to one side:y_{n+1}y_n + 12y_{n+1} + 5y_n - y_n² - 16y_n = 0Simplify:y_{n+1}y_n + 12y_{n+1} - y_n² - 11y_n = 0Factor:y_{n+1}(y_n + 12) = y_n² + 11y_nThus,y_{n+1} = (y_n² + 11y_n)/(y_n + 12)Hmm, that's a bit complicated. Maybe I can approximate this for small y_n.Wait, if y_n is small, then y_n² is much smaller than 11y_n, so perhaps for small y_n, y_{n+1} ≈ (11y_n)/(12). That would suggest that y_n decreases by a factor of 11/12 each time, approximately.But let's see. Starting from x₀ = 7, so y₀ = 7 - 5 = 2.So y₀ = 2.Compute y₁:y₁ = (2² + 11*2)/(2 + 12) = (4 + 22)/14 = 26/14 ≈ 1.857y₂ = (1.857² + 11*1.857)/(1.857 + 12) ≈ (3.45 + 20.427)/13.857 ≈ 23.877/13.857 ≈ 1.722y₃ ≈ (1.722² + 11*1.722)/(1.722 + 12) ≈ (2.965 + 18.942)/13.722 ≈ 21.907/13.722 ≈ 1.603y₄ ≈ (1.603² + 11*1.603)/(1.603 + 12) ≈ (2.569 + 17.633)/13.603 ≈ 20.202/13.603 ≈ 1.485y₅ ≈ (1.485² + 11*1.485)/(1.485 + 12) ≈ (2.205 + 16.335)/13.485 ≈ 18.54/13.485 ≈ 1.374y₆ ≈ (1.374² + 11*1.374)/(1.374 + 12) ≈ (1.887 + 15.114)/13.374 ≈ 17.001/13.374 ≈ 1.271y₇ ≈ (1.271² + 11*1.271)/(1.271 + 12) ≈ (1.616 + 13.981)/13.271 ≈ 15.597/13.271 ≈ 1.176y₈ ≈ (1.176² + 11*1.176)/(1.176 + 12) ≈ (1.383 + 12.936)/13.176 ≈ 14.319/13.176 ≈ 1.087y₉ ≈ (1.087² + 11*1.087)/(1.087 + 12) ≈ (1.182 + 11.957)/13.087 ≈ 13.139/13.087 ≈ 1.004y₁₀ ≈ (1.004² + 11*1.004)/(1.004 + 12) ≈ (1.008 + 11.044)/13.004 ≈ 12.052/13.004 ≈ 0.927y₁₁ ≈ (0.927² + 11*0.927)/(0.927 + 12) ≈ (0.86 + 10.197)/12.927 ≈ 11.057/12.927 ≈ 0.855y₁₂ ≈ (0.855² + 11*0.855)/(0.855 + 12) ≈ (0.731 + 9.405)/12.855 ≈ 10.136/12.855 ≈ 0.789y₁₃ ≈ (0.789² + 11*0.789)/(0.789 + 12) ≈ (0.622 + 8.679)/12.789 ≈ 9.301/12.789 ≈ 0.727y₁₄ ≈ (0.727² + 11*0.727)/(0.727 + 12) ≈ (0.528 + 7.997)/12.727 ≈ 8.525/12.727 ≈ 0.669y₁₅ ≈ (0.669² + 11*0.669)/(0.669 + 12) ≈ (0.447 + 7.359)/12.669 ≈ 7.806/12.669 ≈ 0.616y₁₆ ≈ (0.616² + 11*0.616)/(0.616 + 12) ≈ (0.379 + 6.776)/12.616 ≈ 7.155/12.616 ≈ 0.567y₁₇ ≈ (0.567² + 11*0.567)/(0.567 + 12) ≈ (0.321 + 6.237)/12.567 ≈ 6.558/12.567 ≈ 0.522y₁₈ ≈ (0.522² + 11*0.522)/(0.522 + 12) ≈ (0.273 + 5.742)/12.522 ≈ 6.015/12.522 ≈ 0.480y₁₉ ≈ (0.480² + 11*0.480)/(0.480 + 12) ≈ (0.230 + 5.280)/12.480 ≈ 5.510/12.480 ≈ 0.441y₂₀ ≈ (0.441² + 11*0.441)/(0.441 + 12) ≈ (0.194 + 4.851)/12.441 ≈ 5.045/12.441 ≈ 0.406Hmm, it's still decreasing, but the rate is slowing down. Maybe I need a different approach.Wait, earlier I thought that for small y_n, y_{n+1} ≈ (11/12)y_n. If that's the case, then y_n would decrease exponentially with a base of 11/12 each time. So, starting from y₀ = 2, we can model y_n ≈ 2*(11/12)^n.We need y_n ≤ 1/(2²⁵). So,2*(11/12)^n ≤ 1/(2²⁵)Divide both sides by 2:(11/12)^n ≤ 1/(2²⁶)Take natural logarithm on both sides:n*ln(11/12) ≤ -26*ln(2)Since ln(11/12) is negative, we can divide both sides by it, reversing the inequality:n ≥ (26*ln(2))/ln(12/11)Compute ln(2) ≈ 0.6931ln(12/11) ≈ ln(1.0909) ≈ 0.0862So,n ≥ (26*0.6931)/0.0862 ≈ (17.99)/0.0862 ≈ 208.7So, n ≈ 209. But this is an approximation, assuming y_n decreases by 11/12 each time. However, from the earlier computations, the decrease is faster when y_n is larger and slows down as y_n becomes smaller. So, the actual n needed might be larger than 209.Looking at the answer choices, the intervals are [10,30], [31,90], [91,270], [271,810], [811,∞). Since 209 is between 91 and 270, but my approximation might be underestimating, maybe the actual m is around 270 or higher.Wait, but let me think again. The recursion for y_n is y_{n+1} = (y_n² + 11y_n)/(y_n + 12). For small y_n, this approximates to y_{n+1} ≈ (11y_n)/12, as I did before. But for larger y_n, the quadratic term y_n² becomes more significant. So, maybe the initial decrease is faster, and as y_n becomes smaller, the decrease slows down.Given that, the initial terms decrease quite rapidly, but as y_n approaches zero, the decrease slows down. So, the number of steps needed to get y_n below 1/(2²⁵) might be significantly larger than 209.Alternatively, perhaps I can model this as a difference equation. Let me consider the recursion:y_{n+1} = (y_n² + 11y_n)/(y_n + 12)Let me write this as:y_{n+1} = y_n*(y_n + 11)/(y_n + 12)So, y_{n+1} = y_n*(1 - 1/(y_n + 12))Hmm, that's an interesting form. It suggests that each term is a fraction of the previous term, with the fraction being (1 - 1/(y_n + 12)). So, as y_n decreases, the fraction approaches 1, meaning the decrease slows down.Alternatively, for small y_n, 1/(y_n + 12) ≈ 1/12, so y_{n+1} ≈ y_n*(1 - 1/12) = y_n*(11/12), which is consistent with my earlier approximation.But for larger y_n, say y_n = 2, 1/(y_n + 12) = 1/14 ≈ 0.071, so y_{n+1} ≈ y_n*(1 - 0.071) ≈ y_n*0.929, which is a larger decrease than 11/12.So, the initial terms decrease faster, and as y_n becomes smaller, the decrease slows down. Therefore, the number of steps needed to get y_n below 1/(2²⁵) is somewhere between 200 and maybe 300.Looking at the answer choices, the interval [271,810] is option D. Since my rough estimate was around 209, but considering the initial faster decrease, maybe it's closer to 270 or a bit higher, so option D seems plausible.Alternatively, maybe I can use a better approximation. Let's consider the continuous analog of the recursion. For small y_n, the recursion is approximately y_{n+1} ≈ y_n*(11/12). So, the continuous version would be dy/dn ≈ -y_n*(1 - 11/12) = -y_n*(1/12). This is a differential equation dy/dn = -y/12, whose solution is y(n) = y₀*e^{-n/12}.We want y(n) ≤ 1/(2²⁵). So,2*e^{-n/12} ≤ 1/(2²⁵)Divide both sides by 2:e^{-n/12} ≤ 1/(2²⁶)Take natural log:-n/12 ≤ -26*ln(2)Multiply both sides by -12 (inequality reverses):n ≥ 12*26*ln(2)Compute:12*26 = 312312*ln(2) ≈ 312*0.6931 ≈ 216.3So, n ≈ 216.3, which is around 217. Again, this is an approximation, but it's in the same ballpark as before.Given that, and considering the initial faster decrease, the actual m might be a bit higher, say around 270, which is in the interval [271,810].Therefore, I think the answer is D.

💡First, I need to determine the total area of the circular spinner. The formula for the area of a circle is ( A = pi r^2 ), where ( r ) is the radius. Given that the radius is 15 cm, the total area is ( pi times 15^2 = 225pi ) square centimeters.Next, I know that the probability of winning on one spin is ( frac{1}{3} ). This probability represents the ratio of the area of the WIN sector to the total area of the spinner. Therefore, I can set up the equation ( frac{1}{3} = frac{text{area of the WIN sector}}{225pi} ).To find the area of the WIN sector, I multiply both sides of the equation by ( 225pi ), resulting in ( text{area of the WIN sector} = frac{1}{3} times 225pi = 75pi ) square centimeters.Thus, the area of the WIN sector is ( 75pi ) square centimeters.

💡Alright, let's tackle this problem step by step. So, we have twice as many children as adults around a table. Each person's age is a positive integer greater than 1. The sum of the adults' ages is 156. The mean age of the children is 80% less than the mean age of the whole group. We need to find the sum of the ages of the children.First, let's break down the information:1. Let the number of adults be ( n ). Then, the number of children is ( 2n ) because there are twice as many children as adults.2. The sum of the ages of the adults is 156. So, if we denote the sum of the ages of the children as ( T ), then the total sum of ages for the entire group is ( 156 + T ).3. The mean age of the children is 80% less than the mean age of the whole group. That means the children's mean age is 20% of the whole group's mean age.Okay, let's translate this into mathematical expressions.First, the total number of people around the table is ( n + 2n = 3n ).The mean age of the whole group is the total sum of ages divided by the total number of people:[text{Mean age of the group} = frac{156 + T}{3n}]The mean age of the children is:[text{Mean age of children} = frac{T}{2n}]According to the problem, the mean age of the children is 80% less than the mean age of the whole group. That means:[frac{T}{2n} = 0.2 times frac{156 + T}{3n}]Let me explain why it's 0.2. If something is 80% less, it means it's 20% of the original. So, 80% less than the mean age of the group is 20% of that mean age.Now, let's solve this equation step by step.First, write down the equation:[frac{T}{2n} = 0.2 times frac{156 + T}{3n}]Simplify the right side:[0.2 times frac{156 + T}{3n} = frac{0.2(156 + T)}{3n} = frac{31.2 + 0.2T}{3n}]So, now we have:[frac{T}{2n} = frac{31.2 + 0.2T}{3n}]To eliminate the denominators, let's multiply both sides by ( 6n ) (the least common multiple of 2n and 3n):[6n times frac{T}{2n} = 6n times frac{31.2 + 0.2T}{3n}]Simplify both sides:Left side:[6n times frac{T}{2n} = 3T]Right side:[6n times frac{31.2 + 0.2T}{3n} = 2(31.2 + 0.2T) = 62.4 + 0.4T]So now, the equation is:[3T = 62.4 + 0.4T]Subtract ( 0.4T ) from both sides:[3T - 0.4T = 62.4][2.6T = 62.4]Now, solve for ( T ):[T = frac{62.4}{2.6} = 24]So, the sum of the ages of the children is 24.Wait, let me double-check the calculations to make sure I didn't make any mistakes.Starting from:[frac{T}{2n} = 0.2 times frac{156 + T}{3n}]Multiply both sides by ( 6n ):[3T = 0.2 times 2(156 + T)][3T = 0.4(156 + T)][3T = 62.4 + 0.4T][3T - 0.4T = 62.4][2.6T = 62.4][T = frac{62.4}{2.6} = 24]Yes, that seems correct. So, the sum of the ages of the children is 24.

💡Alright, so I have this problem about a sequence {a_n} that satisfies the condition a₁ + a₂ + … + a_n = n - a_n for each natural number n. The problem has two parts: first, I need to prove that the sequence {a_n - 1} is a geometric sequence, and second, I have to find the maximum term of another sequence {b_n} where b_n is defined as (2 - n)(a_n - 1).Okay, let me tackle the first part first. I need to show that {a_n - 1} is a geometric sequence. A geometric sequence is one where each term is a constant multiple of the previous term. So, if I can express a_n - 1 in terms of a_{n-1} - 1 multiplied by some constant ratio, that would do it.Given the condition: a₁ + a₂ + … + a_n = n - a_n.Let me denote S_n as the sum of the first n terms, so S_n = a₁ + a₂ + … + a_n. Then, according to the problem, S_n = n - a_n.But also, S_n = S_{n-1} + a_n, where S_{n-1} is the sum of the first n-1 terms. So, substituting S_n from the problem statement, we have:S_{n-1} + a_n = n - a_n.So, S_{n-1} = n - a_n - a_n = n - 2a_n.But from the problem statement, S_{n-1} should also equal (n - 1) - a_{n-1}, right? Because for n-1, the condition gives S_{n-1} = (n - 1) - a_{n-1}.So, setting these equal:(n - 1) - a_{n-1} = n - 2a_n.Simplify this equation:n - 1 - a_{n-1} = n - 2a_n.Subtract n from both sides:-1 - a_{n-1} = -2a_n.Multiply both sides by -1:1 + a_{n-1} = 2a_n.So, 2a_n = a_{n-1} + 1.Let me rearrange this:a_n = (a_{n-1} + 1)/2.Hmm, that's a recursive formula for a_n in terms of a_{n-1}.Now, I need to find a relationship for a_n - 1. Let's compute a_n - 1:a_n - 1 = (a_{n-1} + 1)/2 - 1 = (a_{n-1} + 1 - 2)/2 = (a_{n-1} - 1)/2.So, a_n - 1 = (a_{n-1} - 1)/2.That's a key equation. It shows that each term a_n - 1 is half of the previous term a_{n-1} - 1. Therefore, the sequence {a_n - 1} is a geometric sequence with common ratio 1/2.Wait, but let me check if the ratio is positive or negative. Let me compute the first few terms to see.Let's compute a₁. For n=1, the condition is a₁ = 1 - a₁, because S₁ = a₁ = 1 - a₁.So, a₁ = 1 - a₁ => 2a₁ = 1 => a₁ = 1/2.So, a₁ - 1 = 1/2 - 1 = -1/2.Then, a₂ - 1 = (a₁ - 1)/2 = (-1/2)/2 = -1/4.Similarly, a₃ - 1 = (a₂ - 1)/2 = (-1/4)/2 = -1/8.So, the sequence {a_n - 1} is: -1/2, -1/4, -1/8, -1/16, ..., which is a geometric sequence with first term -1/2 and common ratio 1/2.So, that proves part (1). The sequence {a_n - 1} is indeed a geometric sequence with ratio 1/2.Now, moving on to part (2). We have to define b_n = (2 - n)(a_n - 1) and find the maximum term of the sequence {b_n}.First, let's express b_n in terms of n. Since we know from part (1) that a_n - 1 is a geometric sequence with first term -1/2 and ratio 1/2, we can write:a_n - 1 = (-1/2) * (1/2)^{n-1} = - (1/2)^n.Wait, let me check that. The general term of a geometric sequence is a_n = a₁ * r^{n-1}. Here, a₁ (for the sequence {a_n - 1}) is -1/2, and r is 1/2. So, a_n - 1 = (-1/2) * (1/2)^{n-1} = (-1/2)^n? Wait, no, let's compute it correctly.Wait, (-1/2) * (1/2)^{n-1} = (-1/2) * (1/2)^{n-1} = (-1) * (1/2)^n. So, a_n - 1 = - (1/2)^n.Yes, that's correct. So, a_n - 1 = - (1/2)^n.Therefore, b_n = (2 - n)(a_n - 1) = (2 - n)(- (1/2)^n) = (n - 2)(1/2)^n.Wait, because (2 - n) is negative when n > 2, so multiplying by - (1/2)^n gives (n - 2)(1/2)^n.So, b_n = (n - 2)(1/2)^n.Now, we need to find the maximum term of the sequence {b_n}.To find the maximum term, we can analyze the behavior of b_n as n increases. Since b_n is defined for n ∈ ℕ*, which is n = 1, 2, 3, ..., we can compute the first few terms and see where the maximum occurs.Alternatively, we can consider the ratio of consecutive terms, b_{n+1}/b_n, and see when this ratio is greater than 1, meaning the sequence is increasing, and when it's less than 1, meaning the sequence is decreasing. The maximum term would occur around the point where the ratio transitions from greater than 1 to less than 1.Let's compute the ratio:b_{n+1}/b_n = [(n + 1 - 2)(1/2)^{n+1}] / [(n - 2)(1/2)^n] = [(n - 1)(1/2)^{n+1}] / [(n - 2)(1/2)^n] = (n - 1)/(n - 2) * (1/2).So, b_{n+1}/b_n = (n - 1)/(2(n - 2)).We can analyze when this ratio is greater than 1, equal to 1, or less than 1.Set (n - 1)/(2(n - 2)) > 1.Multiply both sides by 2(n - 2), assuming n > 2 so that n - 2 > 0:n - 1 > 2(n - 2)n - 1 > 2n - 4Subtract n from both sides:-1 > n - 4Add 4 to both sides:3 > nSo, when n < 3, the ratio is greater than 1, meaning b_{n+1} > b_n.When n = 3:b_{4}/b_3 = (4 - 1)/(2*(4 - 2)) = 3/(2*2) = 3/4 < 1.Wait, but let me compute it correctly.Wait, for n=3, the ratio is (3 - 1)/(2*(3 - 2)) = 2/(2*1) = 1.So, when n=3, the ratio is 1, meaning b_4 = b_3.For n > 3, let's take n=4:b_5/b_4 = (5 - 1)/(2*(5 - 2)) = 4/(2*3) = 4/6 = 2/3 < 1.So, for n=4, the ratio is less than 1, so b_5 < b_4.Similarly, for n=5, the ratio would be (5 - 1)/(2*(5 - 2)) = 4/(2*3) = 2/3 < 1.So, in summary:- For n < 3, b_{n+1} > b_n (increasing)- For n = 3, b_4 = b_3- For n > 3, b_{n+1} < b_n (decreasing)Therefore, the sequence {b_n} increases up to n=3, then remains the same at n=4, and then decreases thereafter.Wait, but let me check the actual values to confirm.Compute b_n for n=1,2,3,4,5,6.Given b_n = (n - 2)(1/2)^n.n=1: (1 - 2)(1/2)^1 = (-1)(1/2) = -1/2n=2: (2 - 2)(1/2)^2 = 0*(1/4) = 0n=3: (3 - 2)(1/2)^3 = 1*(1/8) = 1/8n=4: (4 - 2)(1/2)^4 = 2*(1/16) = 1/8n=5: (5 - 2)(1/2)^5 = 3*(1/32) = 3/32 ≈ 0.09375n=6: (6 - 2)(1/2)^6 = 4*(1/64) = 1/16 ≈ 0.0625Wait, so from n=1 to n=2, b_n goes from -1/2 to 0, which is an increase.From n=2 to n=3, it goes from 0 to 1/8, which is an increase.From n=3 to n=4, it stays at 1/8, so no change.From n=4 to n=5, it decreases from 1/8 to 3/32, which is approximately 0.09375 < 0.125.From n=5 to n=6, it decreases further to 1/16.So, the maximum term is at n=3 and n=4, both equal to 1/8.Therefore, the maximum term of the sequence {b_n} is 1/8, occurring at n=3 and n=4.Wait, but the problem says "find the maximum term", so it's sufficient to state that the maximum is 1/8, achieved at n=3 and n=4.Alternatively, since the problem might expect a single term, but in this case, both n=3 and n=4 give the same maximum value.So, the maximum term is 1/8.