讲解此道题目---**Question Stem:** 若幂级数 $\sum_{n=1}^{\infty} a_n x^n$ 的收敛半径是R, 则幂级数 $\sum_{n=1}^{\infty} a_n (x-2)^n$ 的收敛半径为( ) **Options:** A. 2 R B. 2-R C. 2+RDR D. R

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Let's analyze this power series problem. We have a power series centered at x=0 with radius R, and we need to find the radius of convergence when the series is shifted to center at x=2. The key insight is that the radius of convergence depends only on the coefficients, not the center point. The fundamental principle here is that the radius of convergence depends solely on the coefficients of the power series, not on where it's centered. Both series have identical coefficients a-n, so they must have the same radius of convergence R. Shifting the center from 0 to 2 doesn't change the radius. Let's prove this mathematically. The original series converges when absolute value of x is less than R. For the shifted series, we substitute y equals x minus 2, transforming it back to the same form with identical coefficients. This gives us the condition absolute value of y less than R, which becomes absolute value of x minus 2 less than R. Therefore, the radius remains R. Let's analyze each option systematically. Option A suggests the radius doubles to 2R, but this is incorrect since the coefficients haven't changed. Options B and C imply the radius depends on the center position, which contradicts the fundamental principle. Option D is correct: the radius remains R because only the coefficients determine convergence radius, and they're identical in both series. In conclusion, the answer is D: R. The fundamental principle is that the radius of convergence depends only on the coefficients of the power series, not on where it's centered. Since both series have identical coefficients a-n, they must have the same radius of convergence R. This is a key concept in power series analysis that applies universally.