二項分布
}}p^{k}(1-p)^{n-k}\\&=\sum _{k=1}^{n}n{\frac {(n-1)!}{(n-k)!\,(k-1)!}}pp^{k-1}(1-p)^{n-k}\\&=np\sum _{k=1}^{n}{\frac {(n-1)!}{(n-k)!\,(k-1)!}}p^{k-1}(1-p)^{n-k}\\\end{aligned}}}
ここで、k′ = k − 1とおくと、
= n p ∑ k = 1 n ( n − 1 ) ! ( n − k ) ! ( k − 1 ) ! p k − 1 ( 1 − p ) n − k = n p ∑ k ′ = 0 n − 1 ( n − 1 ) ! ( ( n − 1 ) − k ′ ) ! k ′ ! p k ′ ( 1 − p ) ( n − 1 ) − k ′ = n p ∑ k ′ = 0 n − 1 ( n − 1 k ′ ) p k ′ ( 1 − p ) ( n − 1 ) − k ′ = n p {\displaystyle {\begin{aligned}&{\hphantom {=}}np\sum _{k=1}^{n}{\frac {(n-1)!}{(n-k)!\,(k-1)!}}p^{k-1}(1-p)^{n-k}\\&=np\sum _{k'=0}^{n-1}{\frac {(n-1)!}{{\bigl (}(n-1)-k'{\bigr )}!\,k'!}}p^{k'}(1-p)^{(n-1)-k'}\\&=np\sum _{k'=0}^{n-1}{\binom {n-1}{k'}}p^{k'}(1-p)^{(n-1)-k'}\\&=np\\\end{aligned}}}