2項係数の総和
2項係数の総和
2項係数の総和について次が成り立つ。
\(n\in\mathbb{N}_{0},m\in\mathbb{N}\)とする。
\[ \sum_{j=k}^{n}C\left(j,k\right)=C\left(n+1,k+1\right) \]
2項係数の総和について次が成り立つ。
\(n\in\mathbb{N}_{0},m\in\mathbb{N}\)とする。
(1)
\[ \sum_{k=0}^{n}C\left(n,k\right)=2^{n} \](2)
\[ \sum_{k=0}^{n}P\left(k,m\right)C\left(n,k\right)=P\left(n,m\right)2^{n-m} \](3)
\[ \sum_{k=0}^{n}k^{m}C\left(n,k\right)=n\sum_{k=0}^{n-1}\left(k+1\right)^{m-1}C\left(n-1,k\right) \](4)
\[ \sum_{k=0}^{n}k^{m}C\left(n,k\right)=\sum_{j=0}^{m}S_{2}\left(m,j\right)P\left(n,j\right)2^{n-j} \](5)
\[ \sum_{k=0}^{n}kC\left(n,k\right)=n2^{n-1} \](6)
\[ \sum_{k=0}^{n}k^{2}C\left(n,k\right)=\left(n^{2}+n\right)2^{n-2} \](7)
\(k\in\mathbb{N}_{0}\)とする。\[ \sum_{j=k}^{n}C\left(j,k\right)=C\left(n+1,k+1\right) \]
(1)
\begin{align*} \sum_{k=0}^{n}C\left(n,k\right) & =\sum_{k=0}^{n}C\left(n,k\right)1^{k}1^{n-k}\\ & =\left(1+1\right)^{n}\\ & =2^{n} \end{align*}(2)
\begin{align*} \sum_{k=0}^{n}P\left(k,m\right)C\left(n,k\right) & =\sum_{k=0}^{n}\frac{n!}{\left(k-m\right)!\left(n-k\right)!}\\ & =\frac{n!}{\left(n-m\right)!}\sum_{k=0}^{n}\frac{\left(n-m\right)!}{\left(k-m\right)!\left(n-k\right)!}\\ & =P\left(n,m\right)\sum_{k=0}^{n}C\left(n-m,n-k\right)\\ & =P\left(n,m\right)\sum_{k=0}^{n-m}C\left(n-m,k\right)\\ & =P\left(n,m\right)2^{n-m} \end{align*}(3)
\begin{align*} \sum_{k=0}^{n}k^{m}C\left(n,k\right) & =n\sum_{k=0}^{n}k^{m-1}C\left(n-1,k-1\right)\\ & =n\sum_{k=0}^{n-1}\left(k+1\right)^{m-1}C\left(n-1,k\right) \end{align*}(4)
\begin{align*} \sum_{k=0}^{n}k^{m}C\left(n,k\right) & =\sum_{k=0}^{n}\sum_{j=0}^{m}S_{2}\left(m,j\right)P\left(k,j\right)C\left(n,k\right)\\ & =\sum_{j=0}^{m}S_{2}\left(m,j\right)P\left(n,j\right)2^{n-j} \end{align*}(5)
\begin{align*} \sum_{k=0}^{n}kC\left(n,k\right) & =\sum_{j=0}^{1}S_{2}\left(1,j\right)P\left(n,j\right)2^{n-j}\\ & =S_{2}\left(1,0\right)P\left(n,0\right)2^{n-0}+S_{2}\left(1,1\right)P\left(n,1\right)2^{n-1}\\ & =n2^{n-1} \end{align*}(5)-2
\begin{align*} \sum_{k=0}^{n}kC\left(n,k\right) & =\sum_{k=0}^{n}nC\left(n-1,k-1\right)\\ & =n\sum_{k=1}^{n}C\left(n-1,k-1\right)\\ & =n\sum_{k=0}^{n-1}C\left(n-1,k\right)\\ & =n2^{n-1} \end{align*}(5)-3
\begin{align*} \sum_{k=0}^{n}kC\left(n,k\right) & =\left[\frac{d}{dx}\sum_{k=0}^{n}C\left(n,k\right)x^{k}\right]_{x=1}\\ & =\left[\frac{d}{dx}\left(1+x\right)^{n}\right]_{x=1}\\ & =\left[n\left(1+x\right)^{n-1}\right]_{x=1}\\ & =n2^{n-1} \end{align*}(6)
\begin{align*} \sum_{k=0}^{n}k^{2}C\left(n,k\right) & =\sum_{j=0}^{2}S_{2}\left(2,j\right)P\left(n,j\right)2^{n-j}\\ & =S_{2}\left(2,0\right)P\left(n,0\right)2^{n-0}+S_{2}\left(2,1\right)P\left(n,1\right)2^{n-1}+S_{2}\left(2,2\right)P\left(n,2\right)2^{n-2}\\ & =n2^{n-1}+n\left(n-1\right)2^{n-2}\\ & =\left(n^{2}+n\right)2^{n-2} \end{align*}(6)-2
\begin{align*} \sum_{k=0}^{n}k^{2}C\left(n,k\right) & =\sum_{k=0}^{n}\left\{ k\left(k-1\right)C\left(n,k\right)+kC\left(n,k\right)\right\} \\ & =\sum_{k=0}^{n}\left\{ n\left(n-1\right)C\left(n-2,k-2\right)+nC\left(n-1,k-1\right)\right\} \\ & =n\left(n-1\right)\sum_{k=2}^{n}C\left(n-2,k-2\right)+n\sum_{k=1}^{n}C\left(n-1,k-1\right)\\ & =n\left(n-1\right)\sum_{k=0}^{n-2}C\left(n-2,k-2\right)+n\sum_{k=0}^{n-1}C\left(n-1,k-1\right)\\ & =n\left(n-1\right)2^{n-2}+n2^{n-1}\\ & =\left(n^{2}+n\right)2^{n-2} \end{align*}(6)-3
\begin{align*} \sum_{k=0}^{n}k^{2}C\left(n,k\right) & =\left[\frac{d}{dx}x\frac{d}{dx}\sum_{k=0}^{n}C\left(n,k\right)x^{k}\right]_{x=1}\\ & =\left[\frac{d}{dx}x\frac{d}{dx}\left(1+x\right)^{n}\right]_{x=1}\\ & =n\left[\frac{d}{dx}x\left(1+x\right)^{n-1}\right]_{x=1}\\ & =n\left[\frac{d}{dx}\left\{ \left(1+x\right)^{n}-\left(1+x\right)^{n-1}\right\} \right]_{x=1}\\ & =n\left[n\left(1+x\right)^{n-1}-\left(n-1\right)\left(1+x\right)^{n-2}\right]_{x=1}\\ & =n\left\{ n2^{n-1}-\left(n-1\right)2^{n-2}\right\} \\ & =\left(n^{2}+n\right)2^{n-2} \end{align*}(7)
\begin{align*} \sum_{j=k}^{n}C\left(j,k\right) & =\sum_{j=k}^{n}\left(C\left(j+1,k+1\right)-C\left(j,k+1\right)\right)\cmt{\because\text{パスカルの法則}}\\ & =C\left(n+1,k+1\right)-C\left(k,k+1\right)\\ & =C\left(n+1,k+1\right) \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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2項係数の母関数
\[
\sum_{k=0}^{\infty}C(x+k,k)t^{k}=(1-t)^{-(x+1)}
\]
中央2項係数の通常型母関数
\[
\sum_{k=0}^{\infty}C\left(2k,k\right)z^{k}=\left(1-4z\right)^{-\frac{1}{2}}
\]
2項係数が0になるとき
\[
\forall m,n\in\mathbb{Z},\left(0\leq m<n\right)\lor\left(n<0\leq m\right)\lor\left(m<n<0\right)\Leftrightarrow C\left(m,n\right)=0
\]
2項係数の特殊な積
\[
C(x,t)C(t,y)=C(x,y)C(x-y,x-t)
\]