二项式定理 Binomial Theorem
作者:互联网
组合数
公式一
\[\binom nk=\frac{n!}{k!(n-k)!} \]公式二
\[\binom{n}{k}=\binom{n-1}{k}+\binom{n-1}{k-1} \]公式三
\[\frac1k\binom{n-1}{k-1}=\frac1n\binom nk \]$ \mathbf{Qn1}\quad $证明 \(\displaystyle{\sum_{k=1}^n \binom nk \frac{(-1)^{k+1}}{k}=\sum_{k=1}^n \frac1k }\)
\(\mathbf{Sol1}\)
\[\begin{aligned} \text{LHS}=&\sum_{k=1}^n \binom nk \frac{(-1)^{k+1}}{k}\\ =&\sum_{k=1}^{n-1} \binom nk \frac{(-1)^{k+1}}{k}+\frac{(-1)^{n+1}}{n}\\ =&\sum_{k=1}^{n-1} \binom{n-1}{k} \frac{(-1)^{k+1}}{k}+\sum_{k=1}^{n-1} \binom{n-1}{k-1} \frac{(-1)^{k+1}}{k}+\frac{(-1)^{n+1}}{n}\\ =&\sum_{k=1}^{n-1} \binom{n-1}{k} \frac{(-1)^{k+1}}{k}+\frac1n\sum_{k=1}^{n-1} \binom{n}{k} (-1)^{k+1}+\frac{(-1)^{n+1}}{n}\\ =&\sum_{k=1}^{n-1} \binom{n-1}{k} \frac{(-1)^{k+1}}{k}+\frac1n\sum_{k=1}^n \binom nk(-1)^{k+1}\\ =&\sum_{k=1}^{n-1} \binom{n-1}{k} \frac{(-1)^{k+1}}{k}+\frac1n\\ =&\sum_{k=1}^{n-2} \binom{n-2}{k} \frac{(-1)^{k+1}}{k}+\frac1n+\frac1{n-1}\\ =&\cdots\\ =&\sum_{k=1}^n \frac1k\\ =&\text{RHS} \end{aligned} \]\(\mathbf{Sol2}\)
考虑生成函数
\[S(x)=\sum_{k=1}^n \binom nk \frac{(-1)^{k+1}}{k} x^k,\quad T(x)=\sum_{k=1}^n \frac1k x^k \]\[S'(x)=-\frac1x\sum_{k=1}^n\binom nk(-x)^k=\frac{1-(1-x)^n}{x},\quad T'(x)=\frac{1-x^n}{1-x} \]注意到 \(S(0)=T(0)=0\),而
\[\int_0^1 \frac{1-x^n}{1-x} \text dx=\int_0^1 \frac{1-(1-x)^n}{x} \text dx \]因此
\[S(1)=T(1) \]标签:nk,frac,frac1n,text,sum,Binomial,Theorem,binom,二项式 来源: https://www.cnblogs.com/Arcticus/p/16308154.html