💡Okay, so I have this problem about a geometric sequence. Let me try to understand it step by step. The problem says: The sum of the first n terms of a geometric sequence {a_n} is S_n. For any positive integer k, it holds that a_k equals the limit as n approaches infinity of (S_n - S_k). I need to find the common ratio q.Hmm, let me break this down. First, I know that in a geometric sequence, each term is the previous term multiplied by the common ratio q. So, the nth term a_n is a_1 times q^(n-1). That makes sense.Next, the sum of the first n terms, S_n, for a geometric series is given by the formula S_n = a_1(1 - q^n)/(1 - q), right? That's the formula I remember. So, that's S_n.Now, the problem states that for any positive integer k, a_k equals the limit as n approaches infinity of (S_n - S_k). Let me write that down:a_k = lim_{n→∞} (S_n - S_k)So, substituting the formula for S_n, we have:a_k = lim_{n→∞} [ (a_1(1 - q^n)/(1 - q)) - (a_1(1 - q^k)/(1 - q)) ]Let me simplify that expression inside the limit. Subtracting the two sums:S_n - S_k = [a_1(1 - q^n)/(1 - q)] - [a_1(1 - q^k)/(1 - q)] Factor out a_1/(1 - q):= a_1/(1 - q) [ (1 - q^n) - (1 - q^k) ]Simplify inside the brackets:= a_1/(1 - q) [1 - q^n - 1 + q^k]= a_1/(1 - q) [ -q^n + q^k ]= a_1/(1 - q) [ q^k - q^n ]So, S_n - S_k = a_1(q^k - q^n)/(1 - q)Therefore, the limit as n approaches infinity is:lim_{n→∞} [a_1(q^k - q^n)/(1 - q)]So, a_k = lim_{n→∞} [a_1(q^k - q^n)/(1 - q)]Now, let's analyze this limit. The term q^n as n approaches infinity depends on the value of q. If |q| < 1, then q^n approaches 0. If |q| = 1, it oscillates or stays at 1. If |q| > 1, it goes to infinity. But since we're dealing with a geometric series, and we're taking the limit as n approaches infinity, I think we can assume that |q| < 1 so that the series converges. Otherwise, the limit might not exist or be infinite, which wouldn't make sense here.So, assuming |q| < 1, q^n approaches 0 as n approaches infinity. Therefore, the limit simplifies to:a_k = a_1(q^k - 0)/(1 - q) = a_1 q^k / (1 - q)But wait, from the geometric sequence definition, a_k is also equal to a_1 q^{k - 1}. So, we have:a_1 q^{k - 1} = a_1 q^k / (1 - q)Let me write that equation:a_1 q^{k - 1} = (a_1 q^k) / (1 - q)Hmm, let's solve for q. First, I can cancel a_1 from both sides, assuming a_1 ≠ 0, which it isn't because it's the first term of the sequence.So, q^{k - 1} = q^k / (1 - q)Let me rewrite q^{k - 1} as q^{k} / q:q^{k} / q = q^{k} / (1 - q)So, now, if I have q^{k} / q = q^{k} / (1 - q), I can cancel q^{k} from both sides, assuming q ≠ 0, which it isn't because otherwise all terms after the first would be zero, which doesn't make sense for a geometric sequence.So, 1/q = 1 / (1 - q)Therefore, 1/q = 1 / (1 - q)Cross-multiplying, we get:1 - q = qSo, 1 = 2qTherefore, q = 1/2Okay, so that gives me q = 1/2. Let me check if this makes sense.If q = 1/2, then the sum S_n = a_1(1 - (1/2)^n)/(1 - 1/2) = a_1(1 - (1/2)^n)/(1/2) = 2a_1(1 - (1/2)^n)So, as n approaches infinity, S_n approaches 2a_1, since (1/2)^n approaches 0.Then, S_n - S_k = 2a_1(1 - (1/2)^n) - 2a_1(1 - (1/2)^k) = 2a_1[(1 - (1/2)^n) - (1 - (1/2)^k)] = 2a_1[(1 - 1 + (1/2)^k - (1/2)^n)] = 2a_1[(1/2)^k - (1/2)^n]Taking the limit as n approaches infinity, (1/2)^n approaches 0, so we get 2a_1*(1/2)^k = a_1*(1/2)^{k - 1}But a_k = a_1*(1/2)^{k - 1}, so that matches. Therefore, the condition holds.So, yes, q = 1/2 is the correct common ratio.I think that's the solution. Let me recap:1. Start with the given condition: a_k = lim_{n→∞} (S_n - S_k)2. Express S_n and S_k using the geometric series sum formula.3. Subtract S_k from S_n and take the limit as n approaches infinity.4. Simplify the expression, noting that q^n approaches 0 if |q| < 1.5. Set the resulting expression equal to a_k, which is a_1 q^{k - 1}.6. Solve the equation for q, leading to q = 1/2.7. Verify the solution by plugging q = 1/2 back into the expressions to ensure consistency.Everything seems to check out. So, the common ratio q is 1/2.**Final Answer**The common ratio ( q ) is boxed{frac{1}{2}}.

💡Okay, so I've got this equation here: (2 - x)/(x - 3) = 0. Hmm, I need to find the value of x that makes this equation true. Alright, let's think about how to approach this.First off, I remember that a fraction equals zero when the numerator is zero, provided that the denominator isn't zero because division by zero is undefined. So, my first step should be to set the numerator equal to zero and solve for x. That makes sense.So, the numerator is (2 - x). If I set that equal to zero: 2 - x = 0. Solving for x, I can add x to both sides to get 2 = x, or x = 2. Okay, so x equals 2 is a potential solution.But wait, I need to make sure that this solution doesn't make the denominator zero. The denominator is (x - 3). If I plug x = 2 into the denominator, I get 2 - 3, which is -1. That's not zero, so that's good. That means x = 2 is a valid solution because it doesn't cause any issues with the denominator.Let me double-check my work to make sure I didn't make any mistakes. Starting with the original equation: (2 - x)/(x - 3) = 0. Setting the numerator equal to zero: 2 - x = 0, so x = 2. Plugging x = 2 back into the denominator: 2 - 3 = -1, which is not zero. Everything seems to check out.I guess another way to think about it is to consider the equation as a function. If I have f(x) = (2 - x)/(x - 3), I want to find where f(x) equals zero. Graphically, this would be where the function crosses the x-axis. Since the numerator is zero at x = 2, and the denominator isn't zero there, that's indeed the point where the function equals zero.Just to be thorough, I can also test values around x = 2 to see how the function behaves. For example, if x is slightly less than 2, say x = 1, the numerator becomes 2 - 1 = 1, and the denominator becomes 1 - 3 = -2. So, f(1) = 1/(-2) = -0.5. If x is slightly more than 2, say x = 3, but wait, x can't be 3 because the denominator becomes zero. So, maybe x = 4: numerator is 2 - 4 = -2, denominator is 4 - 3 = 1, so f(4) = -2/1 = -2. Hmm, interesting. So, around x = 2, the function crosses from negative to more negative? Wait, that doesn't seem right.Wait, maybe I should pick a value just above 2, like x = 2.5. Then numerator is 2 - 2.5 = -0.5, denominator is 2.5 - 3 = -0.5. So, f(2.5) = (-0.5)/(-0.5) = 1. Okay, so at x = 2.5, the function is positive. At x = 2, it's zero. At x = 1, it's negative. So, the function crosses from negative to positive as it passes through x = 2. That makes sense because the numerator changes sign at x = 2, and the denominator is negative for x < 3.So, putting it all together, x = 2 is the solution because it makes the numerator zero without affecting the denominator in a bad way. I think I'm confident with this answer now.Just to recap:1. Set numerator equal to zero: 2 - x = 0 → x = 2.2. Check that denominator isn't zero at x = 2: 2 - 3 = -1 ≠ 0.3. Verify by plugging in values around x = 2 to see the function's behavior.Everything lines up, so I'm pretty sure x = 2 is the correct solution.

💡Okay, so I have this problem here: For how many positive integers ( n ) less than or equal to 1000 is ( (cos t + i sin t)^n = cos(nt) + i sin(nt) ) true for all real ( t )? Hmm, interesting. I remember something about Euler's formula and De Moivre's theorem from my trigonometry class. Let me try to recall.First, Euler's formula states that ( e^{it} = cos t + i sin t ). So, if I raise both sides to the power of ( n ), I get ( (e^{it})^n = (cos t + i sin t)^n ). Simplifying the left side, that's ( e^{int} ), which by Euler's formula is ( cos(nt) + i sin(nt) ). So, putting it together, ( (cos t + i sin t)^n = cos(nt) + i sin(nt) ). Wait, that's exactly the equation given in the problem!So, does this mean that the equation is always true for any positive integer ( n )? It seems like it. Let me think again. Is there any condition on ( n ) that I'm missing? The problem specifies that this should hold for all real ( t ). Since Euler's formula and De Moivre's theorem are general for any real number ( t ) and integer ( n ), I don't see any restriction on ( n ) here.But just to be thorough, let me consider what happens if ( n ) is not an integer. For example, if ( n ) is a fraction, then raising ( cos t + i sin t ) to that power might not give a real result or might not follow the same formula. However, the problem specifies that ( n ) is a positive integer, so fractions aren't an issue here.Wait, but what about negative integers? The problem says positive integers, so negatives aren't considered either. So, as long as ( n ) is a positive integer, the equation holds for all real ( t ). Therefore, every positive integer ( n ) from 1 up to 1000 satisfies the equation.Let me double-check with a specific example. Let's take ( n = 2 ). Then, ( (cos t + i sin t)^2 ) should equal ( cos(2t) + i sin(2t) ). Expanding the left side: ( (cos t + i sin t)^2 = cos^2 t - sin^2 t + 2i cos t sin t ). Using the double-angle formulas, ( cos(2t) = cos^2 t - sin^2 t ) and ( sin(2t) = 2 sin t cos t ). So, indeed, ( (cos t + i sin t)^2 = cos(2t) + i sin(2t) ). That checks out.How about ( n = 3 )? Let's compute ( (cos t + i sin t)^3 ). Expanding this, we get ( cos^3 t + 3i cos^2 t sin t - 3 cos t sin^2 t - i sin^3 t ). Grouping the real and imaginary parts: ( (cos^3 t - 3 cos t sin^2 t) + i (3 cos^2 t sin t - sin^3 t) ). Using the triple-angle formulas, ( cos(3t) = cos^3 t - 3 cos t sin^2 t ) and ( sin(3t) = 3 sin t cos^2 t - sin^3 t ). So again, the equation holds.Hmm, so it seems consistent for specific cases. Maybe I should also think about the general case. When we raise ( cos t + i sin t ) to the power of ( n ), using the binomial theorem, we get a sum of terms involving ( cos^{n-k} t ) and ( sin^k t ). However, this expansion is exactly what De Moivre's theorem generalizes, giving ( cos(nt) + i sin(nt) ). So, as long as ( n ) is an integer, the expansion works out to the trigonometric identities for multiple angles.Therefore, since the equation holds for any positive integer ( n ), and the problem asks for how many such ( n ) are less than or equal to 1000, the answer should be all positive integers from 1 to 1000. That means there are 1000 such integers.Wait a second, but let me make sure I'm not missing something. Is there a case where ( n ) being even or odd might cause an issue? For example, does the equation hold for both even and odd ( n )? Let me test ( n = 1 ) and ( n = 2 ) again. For ( n = 1 ), both sides are just ( cos t + i sin t ), so that's trivial. For ( n = 2 ), as I did earlier, it works. For ( n = 3 ), it also works. How about ( n = 4 )?Calculating ( (cos t + i sin t)^4 ), we can either expand it or use De Moivre's theorem. Using De Moivre's theorem, it should be ( cos(4t) + i sin(4t) ). Expanding it, we'd have ( (cos t + i sin t)^2 ) squared, which would give ( (cos(2t) + i sin(2t))^2 ), and then expanding again, we get ( cos(4t) + i sin(4t) ). So, that works too.I think I'm convinced. The equation holds for any positive integer ( n ), regardless of whether it's even or odd, prime or composite. Therefore, all positive integers ( n ) up to 1000 satisfy the equation. So, the number of such integers is 1000.Just to wrap up, the key points are:1. Euler's formula connects complex exponentials to trigonometric functions.2. Raising ( cos t + i sin t ) to the power ( n ) using Euler's formula gives ( cos(nt) + i sin(nt) ).3. This is a direct application of De Moivre's theorem, which is valid for all integers ( n ).4. Testing specific cases like ( n = 2, 3, 4 ) confirms the general result.5. Therefore, all positive integers ( n leq 1000 ) satisfy the equation.So, the final answer is 1000.**Final Answer**The number of valid positive integers ( n ) up to 1000 is boxed{1000}.

💡First, I need to identify the tenths place in the number 45.65001, which is the 6 after the decimal point.Next, I look at the digit in the hundredths place, which is 5, to determine whether to round up or down.According to rounding rules, if the hundredths digit is 5 or greater, I round the tenths digit up by 1.So, I increase the 6 in the tenths place to 7.Finally, I drop all the digits after the tenths place, resulting in 45.7.

💡Okay, so I have this equation here: C_{34}^{2x} = C_{34}^{4x-8}. Hmm, I remember that combinations have some properties, and maybe I can use one of them to solve for x. Let me think... Oh yeah, I think there's a property that says C(n, k) = C(n, n - k). So, that means if two combinations are equal, either their upper indices are the same, or they add up to the lower index. Alright, so applying that property here, I can set up two different cases. First case: The upper indices are equal. So, 2x should be equal to 4x - 8. Let me write that down: 2x = 4x - 8. Hmm, solving for x. If I subtract 2x from both sides, I get 0 = 2x - 8. Then, adding 8 to both sides gives me 2x = 8. Dividing both sides by 2, I find x = 4. Okay, that's one solution.Now, the second case: The sum of the upper indices equals the lower index. So, 2x + (4x - 8) should equal 34. Let me write that: 2x + 4x - 8 = 34. Combining like terms, that's 6x - 8 = 34. Adding 8 to both sides gives me 6x = 42. Dividing both sides by 6, I get x = 7. Wait, so I have two solutions: x = 4 and x = 7. But I should make sure these solutions are valid. Let me check them in the original equation.First, let's plug x = 4 into both sides. The left side is C_{34}^{2*4} = C_{34}^8. The right side is C_{34}^{4*4 - 8} = C_{34}^{16 - 8} = C_{34}^8. So, both sides are equal. That works.Now, let's check x = 7. The left side is C_{34}^{2*7} = C_{34}^{14}. The right side is C_{34}^{4*7 - 8} = C_{34}^{28 - 8} = C_{34}^{20}. Wait a second, C_{34}^{14} and C_{34}^{20}... Are these equal? Hmm, I think they are because of the combination property C(n, k) = C(n, n - k). So, C_{34}^{14} = C_{34}^{34 - 14} = C_{34}^{20}. Yeah, that's correct. So, both solutions are valid.I think that's it. So, the solutions are x = 4 and x = 7.

💡Okay, so I have this problem here: I need to find the value of n such that the coefficient of the x² term in the expansion of (1 - xi)^n is equal to -28. Hmm, let's see. I remember that when you expand a binomial like (a + b)^n, you can use the binomial theorem, which says that each term is of the form C(n, r) * a^(n - r) * b^r, where C(n, r) is the combination of n things taken r at a time. In this case, the binomial is (1 - xi)^n. So, a is 1 and b is -xi. I need to find the coefficient of the x² term, which means I need to look at the term where the power of x is 2. That would correspond to r = 2 in the binomial expansion because b is -xi, and when you raise that to the power of r, you get (-xi)^r, which includes x^r. So, for x², r should be 2.Let me write down the general term for r = 2. It would be C(n, 2) * (1)^(n - 2) * (-xi)^2. Simplifying this, (1)^(n - 2) is just 1, so that term is C(n, 2) * (-xi)^2. Now, (-xi)^2 is (-1)^2 * x² * i². Since (-1)^2 is 1, and i² is -1, so altogether, that becomes 1 * x² * (-1) = -x². So, putting it all together, the term is C(n, 2) * (-x²). Therefore, the coefficient of x² is C(n, 2) * (-1). According to the problem, this coefficient is equal to -28. So, I can set up the equation: C(n, 2) * (-1) = -28. If I multiply both sides by -1, I get C(n, 2) = 28. Now, C(n, 2) is the combination formula, which is n(n - 1)/2. So, substituting that in, I have n(n - 1)/2 = 28. To solve for n, I can multiply both sides by 2 to get rid of the denominator: n(n - 1) = 56. Now, I need to find an integer n such that when I multiply it by one less than itself, I get 56. Let me think about the factors of 56. 56 can be factored as 7 * 8, which are consecutive integers. So, n must be 8 because 8 * 7 = 56. Let me check that: if n = 8, then C(8, 2) = 8 * 7 / 2 = 28, which matches the equation. So, that seems to work. Therefore, the value of n is 8. I think that's the answer. Let me just recap to make sure I didn't make any mistakes. I used the binomial theorem, identified the term with x², calculated its coefficient, set it equal to -28, solved for n, and checked my answer. Everything seems to add up.**Final Answer**The value of ( n ) is boxed{8}.