(1)

\begin{align*} \sum_{k=0}^{\infty}\sum_{j=0}^{k}C(x,j)C(y,k-j)t^{k} & =\sum_{i=0}^{\infty}\sum_{j=0}^{\infty}C(x,j)C(y,i)t^{j+i}\\ & =\sum_{j=0}^{\infty}C(x,j)t^{j}\sum_{i=0}^{\infty}C(y,i)t^{i}\\ & =(t+1)^{x}(t+1)^{y}\\ & =(t+1)^{x+y}\\ & =\sum_{k=0}^{\infty}C(x+y,k)t^{k} \end{align*}

\(t\)の係数を比較すると与式は成り立つ。

(1)-2

\(x,y\)が整数の場合の証明

\begin{align*} \sum_{k=0}^{m+n}\sum_{j=0}^{k}C(m,j)C(n,k-j)x^{k} & =\sum_{i=0}^{m}\sum_{j=0}^{n}C(m,j)C(n,i)x^{j+i}\\ & =\sum_{i=0}^{m}C(n,i)x^{i}\sum_{j=0}^{n}C(m,j)x^{j}\\ & =(x+1)^{m}(x+1)^{n}\\ & =(x+1)^{m+n}\\ & =\sum_{k=0}^{m+n}C(m+n,k)x^{k} \end{align*}

\(x\)の係数を比較すると与式は成り立つ。

(2)

\begin{align*} \sum_{l=0}^{\infty}\sum_{k=0}^{l}C\left(k,m\right)C\left(l-k,n\right)x^{l+1} & =\sum_{j=0}^{\infty}\sum_{k=0}^{\infty}C\left(k,m\right)C\left(j,n\right)x^{k+j+1}\\ & =x\sum_{j=0}^{\infty}C\left(j,n\right)x^{j}\sum_{k=0}^{\infty}C\left(k,m\right)x^{k}\\ & =xx^{n}\left(1-x\right)^{-\left(n+1\right)}x^{m}\left(1-x\right)^{-\left(m+1\right)}\\ & =x^{m+n+1}\left(1-x\right)^{-\left(m+n+2\right)}\\ & =\sum_{l=0}^{\infty}C\left(l,m+n+1\right)x^{l}\\ & =\sum_{l=-1}^{\infty}C\left(l+1,m+n+1\right)x^{l+1}\\ & =\sum_{l=0}^{\infty}C\left(l+1,m+n+1\right)x^{l+1} \end{align*}

\(x^{l+1}\)の係数を比べると、

\[ \sum_{k=0}^{l}C\left(k,m\right)C\left(l-k,n\right)=C\left(l+1,m+n+1\right) \]

となる。