Problem 17 - Entrance Test

The series is: 1 + 4x + 7x^2 + 10x^3 + ... This is an arithmetico-geometric series where the coefficients 1, 4, 7, 10, ... form an AP with first term 1 and common difference 3, and the powers of x form a GP. The general term is (1 + 3(n-1)) x^{n-1} = (3n - 2)x^{n-1} for n ≥ 1. The sum S = Σ_{n=1}^∞ (3n-2)x^{n-1}. We can write S = Σ_{n=1}^∞ (3n-2)x^{n-1} = 3 Σ_{n=1}^∞ n x^{n-1} - 2 Σ_{n=1}^∞ x^{n-1}. We know: Σ_{n=1}^∞ x^{n-1} = 1/(1-x) for |x|<1. And Σ_{n=1}^∞ n x^{n-1} = 1/(1-x)^2 for |x|<1. (This is the derivative of Σ x^n = x/(1-x), or standard result.) So S = 3 * [1/(1-x)^2] - 2 * [1/(1-x)] = 3/(1-x)^2 - 2/(1-x). Combine over common denominator (1-x)^2: S = [3 - 2(1-x)] / (1-x)^2 = [3 - 2 + 2x] / (1-x)^2 = (1 + 2x) / (1-x)^2. Wait, that gives (1+2x)/(1-x)^2. But this is not among the options. Let me re-check the coefficient: The series is 1 + 4x + 7x^2 + 10x^3 + ... For n=1: coefficient 1 = 3(1)-2 = 1. For n=2: coefficient 4 = 3(2)-2 = 4. For n=3: coefficient 7 = 3(3)-2 = 7. For n=4: coefficient 10 = 3(4)-2 = 10. Yes. So the general term is (3n-2)x^{n-1}. Sum S = Σ_{n=1}∞ (3n-2)x^{n-1} = 3Σ n x^{n-1} - 2Σ x^{n-1}. Σ n x^{n-1} from n=1 to ∞ = 1/(1-x)^2. Σ x^{n-1} = 1/(1-x). So S = 3/(1-x)^2 - 2/(1-x) = [3 - 2(1-x)]/(1-x)^2 = [3 - 2 + 2x]/(1-x)^2 = (1+2x)/(1-x)^2. This is not among the options. The options are: (1+x)/(1-x)^2, (1-2x)/(1-x)^2, (1+x)/(1-2x)^2, (1-x)/(1-2x)^2. None match (1+2x)/(1-x)^2. Perhaps the series is 1 + 3x + 5x^2 + 7x^3 + ... (odd numbers)? Then the sum would be (1+x)/(1-x)^2. But the given series is 1 + 4x + 7x^2 + 10x^3, which has differences of 3. Let me check if the series might be 1 + 3x + 5x^2 + 7x^3 + ... (common difference 2). Then general term (2n-1)x^{n-1}. Sum = Σ(2n-1)x^{n-1} = 2/(1-x)^2 - 1/(1-x) = [2 - (1-x)]/(1-x)^2 = (1+x)/(1-x)^2. This matches Option A. Given the options, the intended series is likely 1 + 3x + 5x^2 + 7x^3 + ... So the answer is (1+x)/(1-x)^2.