$$\begin{eqnarray}
k^2-(k-1)^2&=&k^2-(k^2-2k+1)\;\cdots\;左辺を展開する
\\&=&k^2-k^2+2k-1
\\&=&2k-1
\end{eqnarray}$$
\(k\)に\(1\cdots n\)を代入した式を作る． $$\begin{eqnarray} 1^2-(1-1)^2&=&2\cdot 1-1 \\2^2-(2-1)^2&=&2\cdot 2-1 \\3^2-(3-1)^2&=&2\cdot 3-1 \\\vdots& \\n^2-(n-1)^2&=&2\cdot n-1 \end{eqnarray}$$
左辺，右辺を各項ごとに足し合わせる． $$\begin{eqnarray} &\cancel{1^2}&-0^3&=&2\cdot 1&-1 \\&\cancel{2^2}&\cancel{-1^2}&=&2\cdot 2&-1 \\&\cancel{3^2}&\cancel{-2^2}&=&2\cdot 3&-1 \\&&\vdots& \\+)&n^2&\cancel{-(n-1)^2}&=&2\cdot n&-1 \\\hline \\&n^2&-0^2&=&2\sum_{k=1}^n k&-\sum_{k=1}^n 1 \end{eqnarray}$$
\(\sum_{k=1}^n k\)について解く． $$\begin{eqnarray} n^2&=&2\sum_{k=1}^n k-\sum_{k=1}^n 1 \\2\sum_{k=1}^n k&=&n^2+\sum_{k=1}^n 1 \\&=&n^2+n\;\cdots\;\sum_{k=1}^n 1=n \\&=&n(n+1) \\\sum_{k=1}^n k&=&\frac{n(n+1)}{2} \end{eqnarray}$$
$$\begin{eqnarray} k^3-(k-1)^3&=&k^3-(k-1)(k-1)^2\;\cdots\;左辺を展開する \\&=&k^3-(k-1)(k^2-2k+1) \\&=&k^3-\left\{k(k^2-2k+1)-(k^2-2k+1)\right\} \\&=&k^3-\left(k^3-2k^2+k-k^2+2k-1\right) \\&=&k^3-\left(k^3-3k^2+3k-1\right) \\&=&k^3-k^3+3k^2-3k+1 \\&=&3k^2-3k+1 \end{eqnarray}$$
\(k\)に\(1\cdots n\)を代入した式を作る． $$\begin{eqnarray} 1^3-(1-1)^3&=&3\cdot 1^2-3\cdot 1+1 \\2^3-(2-1)^3&=&3\cdot 2^2-3\cdot 2+1 \\3^3-(3-1)^3&=&3\cdot 3^2-3\cdot 3+1 \\&\vdots& \\n^3-(n-1)^3&=&3\cdot n^2-3\cdot n+1 \end{eqnarray}$$
左辺，右辺を各項ごとに足し合わせる． $$\begin{eqnarray} &\cancel{1^3}*&-0^3&=&3\cdot 1^2&-3\cdot 1&+1 \\&\cancel{2^3}&\cancel{-1^3}&=&3\cdot 2^2&-3\cdot 2&+1 \\&\cancel{3^3}&\cancel{-2^3}&=&3\cdot 3^2&-3\cdot 3&+1 \\&&\vdots& \\+)&n^3&\cancel{-(n-1)^3}&=&3\cdot n^2&-3\cdot n&+1 \\\hline \\&n^3&-0^3&=&3\sum_{k=1}^n k^2&-3\sum_{k=1}^n k&+\sum_{k=1}^n 1 \end{eqnarray}$$
\(\sum_{k=1}^n k^2\)について解く． $$\begin{eqnarray} n^3&=&3\sum_{k=1}^n k^2-3\sum_{k=1}^n k+\sum_{k=1}^n 1 \\3\sum_{k=1}^n k^2&=&n^3+3\sum_{k=1}^n k-\sum_{k=1}^n 1 \\&=&n^3+3\frac{n(n+1)}{2}-n\;\cdots\;\sum_{k=1}^n k=\frac{n(n+1)}{2},\;\sum_{k=1}^n 1=n \\&=&n^3+\frac{3}{2}n^2+\frac{3}{2}n-n \\&=&n^3+\frac{3}{2}n^2+\frac{1}{2}n \\&=&\frac{\left(2n^3+3n^2+n\right)}{2} \\&=&\frac{n\left(2n^2+3n+1\right)}{2} \\&=&\frac{n\left(n+1\right)\left(2n+1\right)}{2} \\\sum_{k=1}^n k^2&=&\frac{n\left(n+1\right)\left(2n+1\right)}{6} \end{eqnarray}$$
$$\begin{eqnarray} k^4-(k-1)^4&=&k^4-(k-1)^2(k-1)^2\;\cdots\;左辺を展開する \\&=&k^4-\left\{(k^2-2k+1)(k^2-2k+1)\right\} \\&=&k^4-\left\{k^2(k^2-2k+1)-2k(k^2-2k+1)+(k^2-2k+1)\right\} \\&=&k^4-\left(k^4-2k^3+k^2-2k^3+4k^2-2k+k^2-2k+1\right) \\&=&k^4-\left(k^4-4k^3+6k^2-4k+1\right) \\&=&k^4-k^4+4k^3-6k^2+4k-1 \\&=&4k^3-6k^2+4k-1 \end{eqnarray}$$
\(k\)に\(1\cdots n\)を代入した式を作る． $$\begin{eqnarray} 1^4-(1-1)^4&=&4\cdot1^3&-6\cdot1^2&+4\cdot1&-1 \\2^4-(2-1)^4&=&4\cdot2^3&-6\cdot2^2&+4\cdot2&-1 \\3^4-(3-1)^4&=&4\cdot3^3&-6\cdot3^2&+4\cdot3&-1 \\&\vdots& \\n^4-(n-1)^4&=&4\cdot n^3&-6\cdot n^2&+4\cdot n&-1 \end{eqnarray}$$
左辺，右辺を各項ごとに足し合わせる． $$\begin{eqnarray} &\cancel{1^4}&-0^4&=&4\cdot1^3&-6\cdot1^2&+4\cdot1&-1 \\&\cancel{2^4}&\cancel{-1^4}&=&4\cdot2^3&-6\cdot2^2&+4\cdot2&-1 \\&\cancel{3^4}&\cancel{-2^4}&=&4\cdot3^3&-6\cdot3^2&+4\cdot3&-1 \\&&&\vdots& \\+)&n^4&\cancel{-(n-1)^4}&=&4\cdot n^3&-6\cdot n^2&+4\cdot n&-1 \\\hline \\&n^4&-0^4&=&4\sum_{k=1}^n k^3&-6\sum_{k=1}^n k^2&+4\sum_{k=1}^n k&-\sum_{k=1}^n 1 \end{eqnarray}$$
\(\sum_{k=1}^n k^3\)について解く． $$\begin{eqnarray} n^4&=&4\sum_{k=1}^n k^3-6\sum_{k=1}^n k^2+4\sum_{k=1}^n k-\sum_{k=1}^n 1 \\4\sum_{k=1}^n k^3&=&n^4+6\sum_{k=1}^n k^2-4\sum_{k=1}^n k+\sum_{k=1}^n 1 \\&=&n^4+6\frac{n(n+1)(2n+1)}{6}-4\frac{n(n+1)}{2}+n\;\cdots\;\sum_{k=1}^n k^2=\frac{n(n+1)(2n+1)}{6},\sum_{k=1}^n k=\frac{n(n+1)}{2},\;\sum_{k=1}^n 1=n \\&=&n^4+n(n+1)(2n+1)-2n(n+1)+n \\&=&n^4+(n^2+n)(2n+1)-2n^2-2n+n \\&=&n^4+n^2(2n+1)+n(2n+1)-2n^2-2n+n \\&=&n^4+2n^3+n^2+2n^2+n-2n^2-2n+n \\&=&n^4+2n^3+n^2 \\&=&n^2(n^2+2n+1) \\&=&n^2(n+1)^2 \\\sum_{k=1}^n k^3&=&\frac{n^2(n+1)^2}{4}=\left(\frac{n(n+1)}{2}\right)^2=\left(\sum_{k=1}^n k\right)^2 \end{eqnarray}$$
\(k\)に\(1\cdots n\)を代入した式を作る． $$\begin{eqnarray} 1^2-(1-1)^2&=&2\cdot 1-1 \\2^2-(2-1)^2&=&2\cdot 2-1 \\3^2-(3-1)^2&=&2\cdot 3-1 \\\vdots& \\n^2-(n-1)^2&=&2\cdot n-1 \end{eqnarray}$$
左辺，右辺を各項ごとに足し合わせる． $$\begin{eqnarray} &\cancel{1^2}&-0^3&=&2\cdot 1&-1 \\&\cancel{2^2}&\cancel{-1^2}&=&2\cdot 2&-1 \\&\cancel{3^2}&\cancel{-2^2}&=&2\cdot 3&-1 \\&&\vdots& \\+)&n^2&\cancel{-(n-1)^2}&=&2\cdot n&-1 \\\hline \\&n^2&-0^2&=&2\sum_{k=1}^n k&-\sum_{k=1}^n 1 \end{eqnarray}$$
\(\sum_{k=1}^n k\)について解く． $$\begin{eqnarray} n^2&=&2\sum_{k=1}^n k-\sum_{k=1}^n 1 \\2\sum_{k=1}^n k&=&n^2+\sum_{k=1}^n 1 \\&=&n^2+n\;\cdots\;\sum_{k=1}^n 1=n \\&=&n(n+1) \\\sum_{k=1}^n k&=&\frac{n(n+1)}{2} \end{eqnarray}$$
$$\begin{eqnarray} k^3-(k-1)^3&=&k^3-(k-1)(k-1)^2\;\cdots\;左辺を展開する \\&=&k^3-(k-1)(k^2-2k+1) \\&=&k^3-\left\{k(k^2-2k+1)-(k^2-2k+1)\right\} \\&=&k^3-\left(k^3-2k^2+k-k^2+2k-1\right) \\&=&k^3-\left(k^3-3k^2+3k-1\right) \\&=&k^3-k^3+3k^2-3k+1 \\&=&3k^2-3k+1 \end{eqnarray}$$
\(k\)に\(1\cdots n\)を代入した式を作る． $$\begin{eqnarray} 1^3-(1-1)^3&=&3\cdot 1^2-3\cdot 1+1 \\2^3-(2-1)^3&=&3\cdot 2^2-3\cdot 2+1 \\3^3-(3-1)^3&=&3\cdot 3^2-3\cdot 3+1 \\&\vdots& \\n^3-(n-1)^3&=&3\cdot n^2-3\cdot n+1 \end{eqnarray}$$
左辺，右辺を各項ごとに足し合わせる． $$\begin{eqnarray} &\cancel{1^3}*&-0^3&=&3\cdot 1^2&-3\cdot 1&+1 \\&\cancel{2^3}&\cancel{-1^3}&=&3\cdot 2^2&-3\cdot 2&+1 \\&\cancel{3^3}&\cancel{-2^3}&=&3\cdot 3^2&-3\cdot 3&+1 \\&&\vdots& \\+)&n^3&\cancel{-(n-1)^3}&=&3\cdot n^2&-3\cdot n&+1 \\\hline \\&n^3&-0^3&=&3\sum_{k=1}^n k^2&-3\sum_{k=1}^n k&+\sum_{k=1}^n 1 \end{eqnarray}$$
\(\sum_{k=1}^n k^2\)について解く． $$\begin{eqnarray} n^3&=&3\sum_{k=1}^n k^2-3\sum_{k=1}^n k+\sum_{k=1}^n 1 \\3\sum_{k=1}^n k^2&=&n^3+3\sum_{k=1}^n k-\sum_{k=1}^n 1 \\&=&n^3+3\frac{n(n+1)}{2}-n\;\cdots\;\sum_{k=1}^n k=\frac{n(n+1)}{2},\;\sum_{k=1}^n 1=n \\&=&n^3+\frac{3}{2}n^2+\frac{3}{2}n-n \\&=&n^3+\frac{3}{2}n^2+\frac{1}{2}n \\&=&\frac{\left(2n^3+3n^2+n\right)}{2} \\&=&\frac{n\left(2n^2+3n+1\right)}{2} \\&=&\frac{n\left(n+1\right)\left(2n+1\right)}{2} \\\sum_{k=1}^n k^2&=&\frac{n\left(n+1\right)\left(2n+1\right)}{6} \end{eqnarray}$$
$$\begin{eqnarray} k^4-(k-1)^4&=&k^4-(k-1)^2(k-1)^2\;\cdots\;左辺を展開する \\&=&k^4-\left\{(k^2-2k+1)(k^2-2k+1)\right\} \\&=&k^4-\left\{k^2(k^2-2k+1)-2k(k^2-2k+1)+(k^2-2k+1)\right\} \\&=&k^4-\left(k^4-2k^3+k^2-2k^3+4k^2-2k+k^2-2k+1\right) \\&=&k^4-\left(k^4-4k^3+6k^2-4k+1\right) \\&=&k^4-k^4+4k^3-6k^2+4k-1 \\&=&4k^3-6k^2+4k-1 \end{eqnarray}$$
\(k\)に\(1\cdots n\)を代入した式を作る． $$\begin{eqnarray} 1^4-(1-1)^4&=&4\cdot1^3&-6\cdot1^2&+4\cdot1&-1 \\2^4-(2-1)^4&=&4\cdot2^3&-6\cdot2^2&+4\cdot2&-1 \\3^4-(3-1)^4&=&4\cdot3^3&-6\cdot3^2&+4\cdot3&-1 \\&\vdots& \\n^4-(n-1)^4&=&4\cdot n^3&-6\cdot n^2&+4\cdot n&-1 \end{eqnarray}$$
左辺，右辺を各項ごとに足し合わせる． $$\begin{eqnarray} &\cancel{1^4}&-0^4&=&4\cdot1^3&-6\cdot1^2&+4\cdot1&-1 \\&\cancel{2^4}&\cancel{-1^4}&=&4\cdot2^3&-6\cdot2^2&+4\cdot2&-1 \\&\cancel{3^4}&\cancel{-2^4}&=&4\cdot3^3&-6\cdot3^2&+4\cdot3&-1 \\&&&\vdots& \\+)&n^4&\cancel{-(n-1)^4}&=&4\cdot n^3&-6\cdot n^2&+4\cdot n&-1 \\\hline \\&n^4&-0^4&=&4\sum_{k=1}^n k^3&-6\sum_{k=1}^n k^2&+4\sum_{k=1}^n k&-\sum_{k=1}^n 1 \end{eqnarray}$$
\(\sum_{k=1}^n k^3\)について解く． $$\begin{eqnarray} n^4&=&4\sum_{k=1}^n k^3-6\sum_{k=1}^n k^2+4\sum_{k=1}^n k-\sum_{k=1}^n 1 \\4\sum_{k=1}^n k^3&=&n^4+6\sum_{k=1}^n k^2-4\sum_{k=1}^n k+\sum_{k=1}^n 1 \\&=&n^4+6\frac{n(n+1)(2n+1)}{6}-4\frac{n(n+1)}{2}+n\;\cdots\;\sum_{k=1}^n k^2=\frac{n(n+1)(2n+1)}{6},\sum_{k=1}^n k=\frac{n(n+1)}{2},\;\sum_{k=1}^n 1=n \\&=&n^4+n(n+1)(2n+1)-2n(n+1)+n \\&=&n^4+(n^2+n)(2n+1)-2n^2-2n+n \\&=&n^4+n^2(2n+1)+n(2n+1)-2n^2-2n+n \\&=&n^4+2n^3+n^2+2n^2+n-2n^2-2n+n \\&=&n^4+2n^3+n^2 \\&=&n^2(n^2+2n+1) \\&=&n^2(n+1)^2 \\\sum_{k=1}^n k^3&=&\frac{n^2(n+1)^2}{4}=\left(\frac{n(n+1)}{2}\right)^2=\left(\sum_{k=1}^n k\right)^2 \end{eqnarray}$$
