$\sum _{k=1}^n{k}^3$の計算式

上式を，${\displaystyle \sum _{k=1}^n{k}^3=\frac{1}{4}\left\{{\left(n+1\right)}^4-1-6\sum _{k=1}^n{k}^2-4\sum _{k=1}^{n}k-n\right\}}$ と整理し，右辺に${\displaystyle \sum _{k=1}^{n}k=\frac{1}{2}n\left(n+1\right)}$ （参照）， ${\displaystyle \sum _{k=1}^n{k}^2=\frac{1}{6}n\left(n+1\right)\left(2n+1\right)}$ （参照）をそれぞれ代入する．

${\displaystyle {\displaystyle \sum _{k=1}^n{k}^3}}$${\displaystyle =\frac{1}{4}\left\{{\left(n+1\right)}^4-1-6\sum _{k=1}^n{k}^2-4\sum _{k=1}^{n}k-n\right\}}$

${\displaystyle {\displaystyle {\displaystyle =\frac{1}{4}\left\{{\left(n+1\right)}^4-6\sum _{k=1}^n{k}^2-4\sum _{k=1}^{n}k-n-1\right\}}}}$

${\displaystyle {\displaystyle =\frac{1}{4}\left\{{\left(n+1\right)}^4-n\left(n+1\right)\left(2n+1\right)-2n\left(n+1\right)-\left(n+1\right)\right\}}}$

${\displaystyle {\displaystyle =\frac{1}{4}\left(n+1\right)\left\{{\left(n+1\right)}^3-n\left(2n+1\right)-2n-1\right\}}}$

${\displaystyle {\displaystyle =\frac{1}{4}\left(n+1\right)\left\{\left(n^3+3n^2+3n+1\right)-\left(2n^2+n\right)-2n-1\right\}}}$

${\displaystyle {\displaystyle =\frac{1}{4}\left(n+1\right)\left\{n^3+\left(3-2\right)n^2+\left(3-1-2\right)n+\left(1-1\right)\right\}}}$

${\displaystyle {\displaystyle =\frac{1}{4}\left(n+1\right)\left(n^3+n^2\right)}}$

${\displaystyle {\displaystyle =\frac{1}{4}\left(n^4+n^3+n^3+n^2\right)}}$

${\displaystyle {\displaystyle =\frac{1}{4}\left(n^4+2n^3+n^2\right)}}$

${\displaystyle {\displaystyle =\frac{1}{4}{n}^2\left(n^2+2n+1\right)}}$

${\displaystyle {\displaystyle ={\left\{\frac{1}{2}n\left(n+1\right)\right\}}^2}}$