Question

请问这道题幂级数求和怎么做

Answer 1

记号有点不习惯, 按我熟悉的写, 结果是: ∑{0 ≤ k} C(k+m,k)·x^k = 1/(1-x)^(m+1).
证明可以对m用数学归纳法.
m = 0时1/(1-x) = ∑{0 ≤ k} x^k, 结论成立.
假设m = n时结论成立, 即∑{0 ≤ k} C(k+n,k)·x^k = 1/(1-x)^(n+1).
两边对x求导得∑{1 ≤ k} k·C(k+n,k)·x^(k-1) = (n+1)/(1-x)^(n+2).
而k·C(k+n,k) = k·(k+n)!/(k!·n!) = (n+1)·(k+n)!/((k-1)!·(n+1)!) = (n+1)·C(k+n,k-1).
故∑{1 ≤ k} k·C(k+n,k)·x^(k-1) = (n+1)·∑{1 ≤ k} C(k+n,k-1)·x^(k-1) = (n+1)·∑{0 ≤ k} C(k+n+1,k)·x^k.
即得∑{0 ≤ k} C(k+n+1,k)·x^k = 1/(1-x)^(n+2), m = n+1时结论成立.
由数学归纳法原理, 结论对任意自然数m成立.