(1)

\begin{align*} \sum_{k=0}^{\infty}\sum_{j=0}^{k}C\left(x,j\right)C\left(y,k-j\right)t^{k} & =\sum_{i=0}^{\infty}\sum_{j=0}^{\infty}C\left(x,j\right)C\left(y,i\right)t^{j+i}\\ & =\sum_{j=0}^{\infty}C\left(x,j\right)t^{j}\sum_{i=0}^{\infty}C\left(y,i\right)t^{i}\\ & =\left(t+1\right)^{x}\left(t+1\right)^{y}\\ & =\left(t+1\right)^{x+y}\\ & =\sum_{k=0}^{\infty}C\left(x+y,k\right)t^{k} \end{align*} \(t\)の係数を比較すると与式は成り立つ。

(1)-2

超幾何関数と超幾何定理を使う。
\begin{align*} \sum_{j=0}^{k}C\left(x,j\right)C\left(y,k-j\right) & =\sum_{j=0}^{k}\frac{P\left(x,j\right)P\left(y,k-j\right)}{j!\left(k-j\right)!}\\ & =\sum_{j=0}^{k}\frac{P\left(x,j\right)}{j!\left(k-j\right)!Q\left(y+1,j-k\right)}\\ & =\sum_{j=0}^{k}\frac{P\left(x,j\right)}{j!\left(k-j\right)!Q\left(y+1,-k\right)Q\left(y-k+1,j\right)}\\ & =\frac{1}{k!Q\left(y+1,-k\right)}\sum_{j=0}^{k}\frac{k!P\left(x,j\right)}{j!\left(k-j\right)!Q\left(y-k+1,j\right)}\\ & =\frac{1}{k!Q\left(y+1,-k\right)}\sum_{j=0}^{k}\frac{P\left(k,j\right)P\left(x,j\right)}{j!Q\left(y-k+1,j\right)}\\ & =\frac{1}{k!Q\left(y+1,-k\right)}\sum_{j=0}^{k}\frac{\left(-1\right)^{j}P\left(k,j\right)\left(-1\right)^{j}P\left(x,j\right)}{j!Q\left(y-k+1,j\right)}\\ & =\frac{1}{k!Q\left(y+1,-k\right)}\sum_{j=0}^{k}\frac{Q\left(-k,j\right)Q\left(-x,j\right)}{j!Q\left(y-k+1,j\right)}\\ & =\frac{1}{k!Q\left(y+1,-k\right)}F\left(-k,-x;y-k+1;1\right)\\ & =\frac{1}{k!Q\left(y+1,-k\right)}\cdot\frac{\Gamma\left(y-k+1\right)\Gamma\left(y-k+1+k+x\right)}{\Gamma\left(y-k+1+k\right)\Gamma\left(y-k+1+x\right)}\cmt{\text{超幾何定理}}\\ & =\frac{1}{k!Q\left(y+1,-k\right)}\cdot\frac{\Gamma\left(y-k+1\right)\Gamma\left(y+1+x\right)}{\Gamma\left(y+1\right)\Gamma\left(y-k+1+x\right)}\\ & =\frac{1}{k!Q\left(y+1,-k\right)}\cdot\frac{\Gamma\left(y\right)Q\left(y,1-k\right)\Gamma\left(y+x\right)\left(y+x\right)}{\Gamma\left(y\right)y\Gamma\left(y+x\right)Q\left(y+x,1-k\right)}\\ & =\frac{1}{k!Q\left(y,1-k\right)}\cdot\frac{Q\left(y,1-k\right)Q\left(y+x,1\right)}{Q\left(y+x,1-k\right)}\\ & =\frac{\left(y+x\right)}{k!Q\left(y+x,1-k\right)}\\ & =\frac{\left(y+x\right)P\left(y+x-1,k-1\right)}{k!}\\ & =\frac{P\left(y+x,k\right)}{k!}\\ & =C\left(x+y,k\right) \end{align*}

(1)-3

\(x,y\)が整数の場合の証明
\(x+y\)個の対象から\(n\)個を選ぶ方法は\(C\left(x+y,n\right)\)通りである。
これは\(x\)個と\(y\)個に分けて\(x\)個の中から\(k\)個と\(y\)個の中から\(n-k\)個を選び\(k\)について0から\(n\)まで総和をとっても同じである。
すなわち\(\sum_{k=0}^{n}C\left(x,k\right))C\left(y,n-k\right)=C\left(x+y,k\right)\)となる。

(1)-4

\(x,y\)が整数の場合の証明
\begin{align*} \sum_{k=0}^{m+n}\sum_{j=0}^{k}C\left(m,j\right)C\left(n,k-j\right)x^{k} & =\sum_{i=0}^{m}\sum_{j=0}^{n}C\left(m,j\right)C\left(n,i\right)x^{j+i}\\ & =\sum_{i=0}^{m}C\left(n,i\right)x^{i}\sum_{j=0}^{n}C\left(m,j\right)x^{j}\\ & =\left(x+1\right)^{m}\left(x+1\right)^{n}\\ & =\left(x+1\right)^{m+n}\\ & =\sum_{k=0}^{m+n}C\left(m+n,k\right)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。