# $\sum _{k=1}^{n}{k}^{3}$の計算式

$\sum _{k=1}^{n}{k}^{3}={1}^{3}+{2}^{3}+{3}^{3}+\cdots +{n}^{3}$$={\left\{\frac{n\left(n+1\right)}{2}\right\}}^{2}$

## ■公式の証明

${\left(k+1\right)}^{4}-{k}^{4}=4{k}^{3}+6{k}^{2}+4k+1$ に順に $k=1,2,3,\cdot \cdot \cdot ,n$ を代入し，以下のように縦にそろえて加えると

 ${2}^{4}-{1}^{4}$ $=$ $4·{1}^{3}$ $+6·{1}^{2}$ $+4·1$ $+1$ ${3}^{4}-{2}^{4}$ $=$ $4·{2}^{3}$ $+6·{2}^{2}$ $+4·2$ $+1$ ${4}^{4}-{3}^{4}$ $=$ $4·{3}^{3}$ $+6·{3}^{2}$ $+4·3$ $+1$ $\cdot$ $\cdot$ $\cdot$ $+\right)$ ${\left(n+1\right)}^{4}-{n}^{4}$ $=$ $4·{n}^{3}$ $+6\cdot {n}^{2}$ $+4·n$ $+1$ $\overline{\text{ }\text{ }\text{ }\text{ }{\left(n+1\right)}^{4}-1\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=4\sum _{k=1}^{n}{k}^{3}+6\sum _{k=1}^{n}{k}^{2}+4\sum _{k=1}^{n}k+n}$

$\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\}$

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

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

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

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

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

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

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

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

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

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

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