1からnまでの三乗の和は，1からnまでの和の二乗

$$\begin{array}{rcl}{k}^2-(k-1{)}^2& =& k^2-(k^2-2k+1)\cdots \mathrm{左}\mathrm{辺}\mathrm{を}\mathrm{展}\mathrm{開}\mathrm{す}\mathrm{る}\\ & =& k^2-k^2+2k-1\\ & =& 2k-1\end{array}$$
$k$に$1\cdots n$を代入した式を作る． $$\begin{array}{rcl}{1}^2-(1-1{)}^2& =& 2\cdot 1-1\\ 2^2-(2-1{)}^2& =& 2\cdot 2-1\\ 3^2-(3-1{)}^2& =& 2\cdot 3-1\\ ⋮& \\ n^2-(n-1{)}^2& =& 2\cdot n-1\end{array}$$
左辺，右辺を各項ごとに足し合わせる． $$\begin{array}{rclrcl}& \cancel{1^2}& -0^3& =& 2\cdot 1& -1\\ & \cancel{2^2}& \cancel{-1^2}& =& 2\cdot 2& -1\\ & \cancel{3^2}& \cancel{-2^2}& =& 2\cdot 3& -1\\ & & ⋮& \\ +)& n^2& \cancel{-(n-1{)}^2}& =& 2\cdot n& -1\\ \\ & n^2& -0^2& =& 2\sum _{k=1}^{n}k& -\sum _{k=1}^{n}1\end{array}$$
$\sum _{k=1}^{n}k$について解く． $$\begin{array}{rcl}{n}^2& =& 2\sum _{k=1}^{n}k-\sum _{k=1}^{n}1\\ 2\sum _{k=1}^{n}k& =& n^2+\sum _{k=1}^{n}1\\ & =& n^2+n\cdots \sum _{k=1}^{n}1=n\\ & =& n(n+1)\\ \sum _{k=1}^{n}k& =& \frac{n(n+1)}{2}\end{array}$$

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$$\begin{array}{rcl}{k}^3-(k-1{)}^3& =& k^3-(k-1)(k-1{)}^2\cdots \mathrm{左}\mathrm{辺}\mathrm{を}\mathrm{展}\mathrm{開}\mathrm{す}\mathrm{る}\\ & =& k^3-(k-1)(k^2-2k+1)\\ & =& k^3-\{k(k^2-2k+1)-(k^2-2k+1)\}\\ & =& k^3-(k^3-2k^2+k-k^2+2k-1)\\ & =& k^3-(k^3-3k^2+3k-1)\\ & =& k^3-k^3+3k^2-3k+1\\ & =& 3k^2-3k+1\end{array}$$
$k$に$1\cdots n$を代入した式を作る． $$\begin{array}{rcl}{1}^3-(1-1{)}^3& =& 3\cdot 1^2-3\cdot 1+1\\ 2^3-(2-1{)}^3& =& 3\cdot 2^2-3\cdot 2+1\\ 3^3-(3-1{)}^3& =& 3\cdot 3^2-3\cdot 3+1\\ & ⋮& \\ n^3-(n-1{)}^3& =& 3\cdot n^2-3\cdot n+1\end{array}$$
左辺，右辺を各項ごとに足し合わせる． $$\begin{array}{rclrclr}& \cancel{1^3}\ast & -0^3& =& 3\cdot 1^2& -3\cdot 1& +1\\ & \cancel{2^3}& \cancel{-1^3}& =& 3\cdot 2^2& -3\cdot 2& +1\\ & \cancel{3^3}& \cancel{-2^3}& =& 3\cdot 3^2& -3\cdot 3& +1\\ & & ⋮& \\ +)& n^3& \cancel{-(n-1{)}^3}& =& 3\cdot n^2& -3\cdot n& +1\\ \\ & n^3& -0^3& =& 3\sum _{k=1}^n{k}^2& -3\sum _{k=1}^{n}k& +\sum _{k=1}^{n}1\end{array}$$
$\sum _{k=1}^n{k}^2$について解く． $$\begin{array}{rcl}{n}^3& =& 3\sum _{k=1}^n{k}^2-3\sum _{k=1}^{n}k+\sum _{k=1}^{n}1\\ 3\sum _{k=1}^n{k}^2& =& n^3+3\sum _{k=1}^{n}k-\sum _{k=1}^{n}1\\ & =& n^3+3\frac{n(n+1)}{2}-n\cdots \sum _{k=1}^{n}k=\frac{n(n+1)}{2},\sum _{k=1}^{n}1=n\\ & =& n^3+\frac{3}{2}{n}^2+\frac{3}{2}n-n\\ & =& n^3+\frac{3}{2}{n}^2+\frac{1}{2}n\\ & =& \frac{(2n^3+3n^2+n)}{2}\\ & =& \frac{n(2n^2+3n+1)}{2}\\ & =& \frac{n(n+1)(2n+1)}{2}\\ \sum _{k=1}^n{k}^2& =& \frac{n(n+1)(2n+1)}{6}\end{array}$$

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$$\begin{array}{rcl}{k}^4-(k-1{)}^4& =& k^4-(k-1{)}^2(k-1{)}^2\cdots \mathrm{左}\mathrm{辺}\mathrm{を}\mathrm{展}\mathrm{開}\mathrm{す}\mathrm{る}\\ & =& k^4-\{(k^2-2k+1)(k^2-2k+1)\}\\ & =& k^4-\{k^2(k^2-2k+1)-2k(k^2-2k+1)+(k^2-2k+1)\}\\ & =& k^4-(k^4-2k^3+k^2-2k^3+4k^2-2k+k^2-2k+1)\\ & =& k^4-(k^4-4k^3+6k^2-4k+1)\\ & =& k^4-k^4+4k^3-6k^2+4k-1\\ & =& 4k^3-6k^2+4k-1\end{array}$$
$k$に$1\cdots n$を代入した式を作る． $$\begin{array}{rclrcl}{1}^4-(1-1{)}^4& =& 4\cdot 1^3& -6\cdot 1^2& +4\cdot 1& -1\\ 2^4-(2-1{)}^4& =& 4\cdot 2^3& -6\cdot 2^2& +4\cdot 2& -1\\ 3^4-(3-1{)}^4& =& 4\cdot 3^3& -6\cdot 3^2& +4\cdot 3& -1\\ & ⋮& \\ n^4-(n-1{)}^4& =& 4\cdot n^3& -6\cdot n^2& +4\cdot n& -1\end{array}$$
左辺，右辺を各項ごとに足し合わせる． $$\begin{array}{rclrclrc}& \cancel{1^4}& -0^4& =& 4\cdot 1^3& -6\cdot 1^2& +4\cdot 1& -1\\ & \cancel{2^4}& \cancel{-1^4}& =& 4\cdot 2^3& -6\cdot 2^2& +4\cdot 2& -1\\ & \cancel{3^4}& \cancel{-2^4}& =& 4\cdot 3^3& -6\cdot 3^2& +4\cdot 3& -1\\ & & & ⋮& \\ +)& n^4& \cancel{-(n-1{)}^4}& =& 4\cdot n^3& -6\cdot n^2& +4\cdot n& -1\\ \\ & n^4& -0^4& =& 4\sum _{k=1}^n{k}^3& -6\sum _{k=1}^n{k}^2& +4\sum _{k=1}^{n}k& -\sum _{k=1}^{n}1\end{array}$$
$\sum _{k=1}^n{k}^3$について解く． $$\begin{array}{rcl}{n}^4& =& 4\sum _{k=1}^n{k}^3-6\sum _{k=1}^n{k}^2+4\sum _{k=1}^{n}k-\sum _{k=1}^{n}1\\ 4\sum _{k=1}^n{k}^3& =& n^4+6\sum _{k=1}^n{k}^2-4\sum _{k=1}^{n}k+\sum _{k=1}^{n}1\\ & =& n^4+6\frac{n(n+1)(2n+1)}{6}-4\frac{n(n+1)}{2}+n\cdots \sum _{k=1}^n{k}^2=\frac{n(n+1)(2n+1)}{6},\sum _{k=1}^{n}k=\frac{n(n+1)}{2},\sum _{k=1}^{n}1=n\\ & =& n^4+n(n+1)(2n+1)-2n(n+1)+n\\ & =& n^4+(n^2+n)(2n+1)-2n^2-2n+n\\ & =& n^4+n^2(2n+1)+n(2n+1)-2n^2-2n+n\\ & =& n^4+2n^3+n^2+2n^2+n-2n^2-2n+n\\ & =& n^4+2n^3+n^2\\ & =& n^2(n^2+2n+1)\\ & =& n^2(n+1{)}^2\\ \sum _{k=1}^n{k}^3& =& \frac{n^2(n+1{)}^2}{4}={\left(\frac{n(n+1)}{2}\right)}^2={\left(\sum _{k=1}^{n}k\right)}^2\end{array}$$