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裂项相消
核心
∑k=1n[F(k)−F(k−1)]=F(n)−F(0)\sum_{k=1}^{n} \bigl[ F(k) - F(k-1) \bigr] = F(n) - F(0)

即相邻项相减后,中间项全部抵消,只剩首尾

常见模式#

分式型#

  1. 1n(n+1)=1n−1n+1\displaystyle \frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1};
  2. 常数倍:1n(n+k)=1k(1n−1n+k)\displaystyle \frac{1}{n(n+k)} = \frac{1}{k}\left(\frac{1}{n} - \frac{1}{n+k}\right);
  3. 三项:1n(n+1)(n+2)=12[1n(n+1)−1(n+1)(n+2)]\displaystyle \frac{1}{n(n+1)(n+2)} = \frac{1}{2}\left[\frac{1}{n(n+1)} - \frac{1}{(n+1)(n+2)}\right]。

根式型(分子有理化)#

基础:(n+1−n)(n+1+n)=1(\sqrt{n+1} - \sqrt{n})(\sqrt{n+1} + \sqrt{n}) = 1,从而

1n+n+1=n+1−n,\frac{1}{\sqrt{n} + \sqrt{n+1}} = \sqrt{n+1} - \sqrt{n},

常数倍:1n+n+k=1k(n+k−n)\displaystyle \frac{1}{\sqrt{n} + \sqrt{n+k}} = \frac{1}{k}\left(\sqrt{n+k} - \sqrt{n}\right)。

指数型#

2n(2n−1)(2n+1−1)=12n−1−12n+1−1.\frac{2^n}{(2^n-1)(2^{n+1}-1)} = \frac{1}{2^n-1} - \frac{1}{2^{n+1}-1}.

阶乘型#

n(n+1)!=(n+1)−1(n+1)!=1n!−1(n+1)!.\frac{n}{(n+1)!} = \frac{(n+1)-1}{(n+1)!} = \frac{1}{n!} - \frac{1}{(n+1)!}.

三角#

tan⁡x−tan⁡y=sin⁡(x−y)cos⁡xcos⁡y.\tan x - \tan y = \frac{\sin(x-y)}{\cos x \cos y}.
裂项相消
https://entropy.lighterever.com/posts/裂项相消/
作者
Lighterever
发布于
2026-09-30
许可协议
CC BY-NC-SA 4.0