对二项式定理求导

标签：ni aligned 定理 nix 求导 binom 二项式 niix sum

\[\begin{aligned} (x+1)^n&=\sum_{i=0}^n\binom nix^i\\ (x+1)^n&=\sum_{i=0}^n\binom nix^i\\ ((x+1)^n)'&=(\sum_{i=0}^n\binom nix^i)'\\ n(x+1)^{n-1}&=\sum_{i=0}^n\binom niix^{i-1}\\ 2^{n-1}n&=\sum_{i=0}^ni\binom ni \end{aligned} \]

\[\begin{aligned} (x+1)^n&=\sum_{i=0}^n\binom nix^i\\ ((x+1)^n)'&=(\sum_{i=0}^n\binom nix^i)'\\ n(x+1)^{n-1}&=\sum_{i=0}^n\binom niix^{i-1}\\ (n(x+1)^{n-1})'&=(\sum_{i=0}^n\binom niix^{i-1})'\\ n(n-1)(x+1)^{n-2}&=\sum_{i=0}^n\binom nii(i-1)x^{i-2}\\ n(n-1)2^{n-2}&=\sum_{i=0}^ni(i-1)\binom ni\\ n(n-1)2^{n-2}+2^{n-1}n&=\sum_{i=0}^ni(i-1)\binom ni+\sum_{i=0}^ni\binom ni\\ n(n-1)2^{n-2}+2\times2^{n-2}n&=\sum_{i=0}^ni^2\binom ni\\ n(n+1)2^{n-2}&=\sum_{i=0}^ni^2\binom ni\\ \end{aligned} \]

标签：ni,aligned,定理,nix,求导,binom,二项式,niix,sum
From： https://www.cnblogs.com/bxjz/p/18007223