2項係数の総和
\(n\in\mathbb{N}_{0},m\in\mathbb{N}\)とする。
(1)
\[ \sum_{k=0}^{n}C(n,k)=2^{n} \](2)
\[ \sum_{k=0}^{n}P(k,m)C(n,k)=P(n,m)2^{n-m} \](3)
\[ \sum_{k=0}^{n}k^{m}C(n,k)=n\sum_{k=0}^{n-1}(k+1)^{m-1}C(n-1,k) \](4)
\[ \sum_{k=0}^{n}k^{m}C(n,k)=\sum_{j=0}^{m}S_{2}(m,j)P(n,j)2^{n-j} \](5)
\[ \sum_{k=0}^{n}kC(n,k)=n2^{n-1} \](6)
\[ \sum_{k=0}^{n}k^{2}C(n,k)=(n^{2}+n)2^{n-2} \](1)
\begin{align*} \sum_{k=0}^{n}C(n,k) & =\sum_{k=0}^{n}C(n,k)1^{k}1^{n-k}\\ & =(1+1)^{n}\\ & =2^{n} \end{align*}(2)
\begin{align*} \sum_{k=0}^{n}P(k,m)C(n,k) & =\sum_{k=0}^{n}\frac{n!}{(k-m)!(n-k)!}\\ & =\frac{n!}{(n-m)!}\sum_{k=0}^{n}\frac{(n-m)!}{(k-m)!(n-k)!}\\ & =P(n,m)\sum_{k=0}^{n}C(n-m,n-k)\\ & =P(n,m)\sum_{k=0}^{n-m}C(n-m,k)\\ & =P(n,m)2^{n-m} \end{align*}(3)
\begin{align*} \sum_{k=0}^{n}k^{m}C(n,k) & =n\sum_{k=0}^{n}k^{m-1}C(n-1,k-1)\\ & =n\sum_{k=0}^{n-1}(k+1)^{m-1}C(n-1,k) \end{align*}(4)
\begin{align*} \sum_{k=0}^{n}k^{m}C(n,k) & =\sum_{k=0}^{n}\sum_{j=0}^{m}S_{2}(m,j)P(k,j)C(n,k)\\ & =\sum_{j=0}^{m}S_{2}(m,j)P(n,j)2^{n-j} \end{align*}(5)
\begin{align*} \sum_{k=0}^{n}kC(n,k) & =\sum_{j=0}^{1}S_{2}(1,j)P(n,j)2^{n-j}\\ & =S_{2}(1,0)P(n,0)2^{n-0}+S_{2}(1,1)P(n,1)2^{n-1}\\ & =n2^{n-1} \end{align*}(5)-2
\begin{align*} \sum_{k=0}^{n}kC(n,k) & =\sum_{k=0}^{n}nC(n-1,k-1)\\ & =n\sum_{k=1}^{n}C(n-1,k-1)\\ & =n\sum_{k=0}^{n-1}C(n-1,k)\\ & =n2^{n-1} \end{align*}(5)-3
\begin{align*} \sum_{k=0}^{n}kC(n,k) & =\left[\frac{d}{dx}\sum_{k=0}^{n}C(n,k)x^{k}\right]_{x=1}\\ & =\left[\frac{d}{dx}(1+x)^{n}\right]_{x=1}\\ & =\left[n(1+x)^{n-1}\right]_{x=1}\\ & =n2^{n-1} \end{align*}(6)
\begin{align*} \sum_{k=0}^{n}k^{2}C(n,k) & =\sum_{j=0}^{2}S_{2}(2,j)P(n,j)2^{n-j}\\ & =S_{2}(2,0)P(n,0)2^{n-0}+S_{2}(2,1)P(n,1)2^{n-1}+S_{2}(2,2)P(n,2)2^{n-2}\\ & =n2^{n-1}+n(n-1)2^{n-2}\\ & =(n^{2}+n)2^{n-2} \end{align*}(6)-2
\begin{align*} \sum_{k=0}^{n}k^{2}C(n,k) & =\sum_{k=0}^{n}\left\{ k(k-1)C(n,k)+kC(n,k)\right\} \\ & =\sum_{k=0}^{n}\left\{ n(n-1)C(n-2,k-2)+nC(n-1,k-1)\right\} \\ & =n(n-1)\sum_{k=2}^{n}C(n-2,k-2)+n\sum_{k=1}^{n}C(n-1,k-1)\\ & =n(n-1)\sum_{k=0}^{n-2}C(n-2,k-2)+n\sum_{k=0}^{n-1}C(n-1,k-1)\\ & =n(n-1)2^{n-2}+n2^{n-1}\\ & =(n^{2}+n)2^{n-2} \end{align*}(6)-3
\begin{align*} \sum_{k=0}^{n}k^{2}C(n,k) & =\left[\frac{d}{dx}x\frac{d}{dx}\sum_{k=0}^{n}C(n,k)x^{k}\right]_{x=1}\\ & =\left[\frac{d}{dx}x\frac{d}{dx}(1+x)^{n}\right]_{x=1}\\ & =n\left[\frac{d}{dx}x(1+x)^{n-1}\right]_{x=1}\\ & =n\left[\frac{d}{dx}\left\{ (1+x)^{n}-(1+x)^{n-1}\right\} \right]_{x=1}\\ & =n\left[n(1+x)^{n-1}-(n-1)(1+x)^{n-2}\right]_{x=1}\\ & =n\left\{ n2^{n-1}-(n-1)2^{n-2}\right\} \\ & =(n^{2}+n)2^{n-2} \end{align*}ページ情報
タイトル | 2項係数の総和 |
URL | https://www.nomuramath.com/io7dswnn/ |
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中央2項係数を含む通常型母関数
\[
\sum_{k=0}^{\infty}\frac{1}{k+1}C\left(2k,k\right)z^{k}=\frac{1}{2z}\left\{ 1-\left(1-4z\right)^{\frac{1}{2}}\right\}
\]
2項係数の半分までの総和
\[
\sum_{k=0}^{n-1}C\left(2n-1,k\right)=2^{2n-2}
\]
2項係数の1項間漸化式
\[
C(x+1,y)=\frac{x+1}{x+1-y}C(x,y)
\]
2項変換と交代2項変換の逆変換
\[
a_{n}=\sum_{k=0}^{n}(-1)^{n-k}C(n,k)b_{k}
\]