Question

高数 这个题目真心不会 求幂级数的和函数

Answer 1

∑{1 ≤ n} (2n-1)·x^(2n-2)/2^n = ∑{0 ≤ n} (2n+1)·x^(2n)/2^(n+1)
= ∑{0 ≤ n} n·x^(2n)/2^n+1/2·∑{0 ≤ n} x^(2n)/2^n
= ∑{0 ≤ n} n·(x²/2)^n+1/2·∑{0 ≤ n} (x²/2)^n.

只要分别求∑{0 ≤ n} x^n与∑{0 ≤ n} n·x^n的和函数.
前者由等比数列求和即知∑{0 ≤ n} x^n = 1/(1-x).
两边求导得∑{1 ≤ n} n·x^(n-1) = 1/(1-x)².
于是∑{0 ≤ n} n·x^n = ∑{1 ≤ n} n·x^n = x/(1-x)².

原式 = (x²/2)/(1-x²/2)²+1/2·1/(1-x²/2) = (x²+2)/(2-x²)².

Answer 2

追问

您用的是软件吧 考试怎么做？

Answer 3