総和
∑ i = 1 n i 5 = 1 12 n 2 ( n + 1 ) 2 ( 2 n 2 + 2 n − 1 ) {\displaystyle \textstyle \sum \limits _{i=1}^{n}i^{5}={\dfrac {1}{12}}n^{2}(n+1)^{2}(2n^{2}+2n-1)}
∑ i = 1 n i ( i + 1 ) ( i + 2 ) ⋯ ( i + k ) = n ( n + 1 ) ( n + 2 ) ⋯ ( n + k + 1 ) k + 2 {\displaystyle \textstyle \sum \limits _{i=1}^{n}{i(i+1)(i+2)\cdots (i+k)}={\dfrac {n(n+1)(n+2)\cdots (n+k+1)}{k+2}}}
∑ i = 0 n i m = ( n + 1 ) m + 1 m + 1 + ∑ k = 1 m B k m − k + 1 ( m k ) ( n + 1 ) m − k + 1 {\displaystyle \textstyle \sum \limits _{i=0}^{n}i^{m}={\dfrac {(n+1)^{m+1}}{m+1}}+\sum \limits _{k=1}^{m}{\dfrac {B_{k}}{m-k+1}}\displaystyle {m \choose k}(n+1)^{m-k+1}} (Bk はベルヌーイ数)