飛び飛びの2項定理
飛び飛びの2項定理
2項定理で飛び飛びのとき次が成り立つ。
2項定理で飛び飛びのとき次が成り立つ。
(1)偶数
\[ \sum_{k=0}^{\infty}C\left(n,2k\right)a^{2k}b^{n-2k}=\frac{1}{2}\left\{ \left(a+b\right)^{n}+\left(-a+b\right)^{n}\right\} \](2)奇数
\[ \sum_{k=0}^{\infty}C\left(n,2k+1\right)a^{2k+1}b^{n-\left(2k+1\right)}=\frac{1}{2}\left\{ \left(a+b\right)^{n}-\left(-a+b\right)^{n}\right\} \](3)一般形
\[ \sum_{k=-\infty}^{\infty}C\left(n,mk+l\right)a^{mk+l}b^{n-\left(mk+l\right)}=\frac{1}{m}\sum_{j=0}^{m-1}\omega_{m}^{-jl}\left(a\omega_{m}^{j}+b\right)^{n} \]\(\omega\)は1の\(n\)乗根、
\[ \omega_{n}=e^{\frac{2\pi}{n}i} \] である。
\[ \omega_{n}=e^{\frac{2\pi}{n}i} \] である。
(1)
\begin{align*} \sum_{k=0}^{\infty}C\left(n,2k\right)a^{2k}b^{n-2k} & =\frac{1}{2}\sum_{k=0}^{\infty}\left(1+\left(-1\right)^{k}\right)C\left(n,k\right)a^{k}b^{n-k}\\ & =\frac{1}{2}\left\{ \sum_{k=0}^{\infty}C\left(n,k\right)a^{k}b^{n-k}+\sum_{k=0}^{\infty}\left(-1\right)^{k}C\left(n,k\right)a^{k}b^{n-k}\right\} \\ & =\frac{1}{2}\left\{ \sum_{k=0}^{\infty}C\left(n,k\right)a^{k}b^{n-k}+\sum_{k=0}^{\infty}C\left(n,k\right)\left(-a\right)^{k}b^{n-k}\right\} \\ & =\frac{1}{2}\left\{ \left(a+b\right)^{n}+\left(-a+b\right)^{n}\right\} \end{align*}(2)
\begin{align*} \sum_{k=0}^{\infty}C\left(n,2k+1\right)a^{2k+1}b^{n-\left(2k+1\right)} & =\sum_{k=0}^{\infty}C\left(n,k\right)a^{k}b^{n-k}-\sum_{k=0}^{\infty}C\left(n,2k\right)a^{2k}b^{n-2k}\\ & =\left(a+b\right)^{n}-\frac{1}{2}\left\{ \left(a+b\right)^{n}+\left(-a+b\right)^{n}\right\} \\ & =\frac{1}{2}\left\{ \left(a+b\right)^{n}-\left(-a+b\right)^{n}\right\} \end{align*}(2)-2
\begin{align*} \sum_{k=0}^{\infty}C\left(n,2k+1\right)a^{2k+1}b^{n-\left(2k+1\right)} & =\frac{1}{2}\sum_{k=0}^{\infty}\left(1-\left(-1\right)^{k}\right)C\left(n,k\right)a^{k}b^{n-k}\\ & =\frac{1}{2}\left\{ \sum_{k=0}^{\infty}C\left(n,k\right)a^{k}b^{n-k}-\sum_{k=0}^{\infty}\left(-1\right)^{k}C\left(n,k\right)a^{k}b^{n-k}\right\} \\ & =\frac{1}{2}\left\{ \sum_{k=0}^{\infty}C\left(n,k\right)a^{k}b^{n-k}-\sum_{k=0}^{\infty}C\left(n,k\right)\left(-a\right)^{k}b^{n-k}\right\} \\ & =\frac{1}{2}\left\{ \left(a+b\right)^{n}-\left(-a+b\right)^{n}\right\} \end{align*}(3)
\[ \sum_{k=0}^{n-1}\left(\omega_{n}^{k}\right)^{m}=n\delta_{0,\mod\left(m,n\right)} \] が成り立つので、\begin{align*} \sum_{k=-\infty}^{\infty}C\left(n,mk+l\right)a^{mk+l}b^{n-\left(mk+l\right)} & =\sum_{k=-\infty}^{\infty}C\left(n,k\right)a^{k}b^{n-k}\delta_{0,\mod\left(k-l,m\right)}\\ & =\sum_{k=-\infty}^{\infty}C\left(n,k\right)a^{k}b^{n-k}\frac{1}{m}\sum_{j=0}^{m-1}\left(\omega_{m}^{j}\right)^{k-l}\\ & =\frac{1}{m}\left(\omega_{m}^{j}\right)^{-l}\sum_{j=0}^{m-1}\sum_{k=-\infty}^{\infty}C\left(n,k\right)\left(a\omega_{m}^{j}\right)^{k}b^{n-k}\\ & =\frac{1}{m}\sum_{j=0}^{m-1}\omega_{m}^{-jl}\left(a\omega_{m}^{j}+b\right)^{n} \end{align*}
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2項係数の微分
\[
\frac{d}{dx}C(x,y) =C(x,y)\left(\psi(1+x)-\psi(1+x-y)\right)
\]
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項係数の逆数の差分
\[
C^{-1}(k+j+1,j+1)=\frac{j+1}{j}\left(C^{-1}(k+j,j)-C^{-1}(k+j+1,j)\right)
\]
2項係数の総和
\[
\sum_{k=0}^{n}P(k,m)C(n,k)=P(n,m)2^{n-m}
\]