数列の総和の公式
本記事では，数列の総和の公式を証明します．
よく教科書や参考書で見られる証明方法だけでなく，数学的帰納法による証明もしてみました．
\sum_{k=1}^n k
公式．
\begin{aligned}
\sum_{k=1}^n k = \frac{1}{2} n (n + 1) \\
\end{aligned}
証明．
\begin{aligned}
\sum_{k=1}^n k = 1 + 2 + \cdots + n \\
\end{aligned}
は交差 1 ，初項 1 ，末項 n ，項数 n の等差数列なので，
\begin{aligned}
\sum_{k=1}^n k &= 1 + 2 + \cdots + n \\
&= \frac{1}{2} n (n + 1) \\
\end{aligned}
\sum_{k=1}^n k^2
公式．
\begin{aligned}
\sum_{k=1}^n k^2 = \frac{1}{6} n (n + 1) (2n + 1) \\
\end{aligned}
証明1．
\begin{aligned}
(k + 1)^3 &= k^3 + 3 k^2 + 3 k + 1 \\
(k + 1)^3 - k^3 &= 3 k^2 + 3 k + 1
\end{aligned}
となることから，
\begin{aligned}
\sum_{k=1}^n \{(k + 1)^3 - k^3 \} &= \sum_{k=1}^n ( 3 k^2 + 3 k + 1) \\
\end{aligned}
が成り立つ．
\begin{aligned}
(左辺) &= \sum_{k=1}^n \{(k + 1)^3 - k^3 \} \\
&= \sum_{k=1}^n (k + 1)^3 - \sum_{k=1}^n k^3 \\
&= (n + 1)^3 - 1^3 \\
&= n^3 + 3 n^2 + 3 n
\end{aligned}
\begin{aligned}
(右辺) &= \sum_{k=1}^n ( 3 k^2 + 3 k + 1) \\
&= 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 + 3 \cdot \frac{1}{2} \cdot n \cdot (n+1) + n \\
&= 3 \sum_{k=1}^n k^2 + \frac{3}{2}n^2 + \frac{5}{2}n
\end{aligned}
したがって，
\begin{aligned}
\sum_{k=1}^n \{(k + 1)^3 - k^3 \} &= \sum_{k=1}^n ( 3 k^2 + 3 k + 1) \\
n^3 + 3 n^2 + 3 n &= 3 \sum_{k=1}^n k^2 + \frac{3}{2}n^2 + \frac{5}{2}n \\
\end{aligned}
となる．
\begin{aligned}
3 \sum_{k=1}^n k^2 &= n^3 + 3 n^2 + 3 n - \frac{3}{2}n^2 - \frac{5}{2}n \\
&= n^3 + \frac{3}{2}n^2 n^2 + \frac{1}{2} n \\
&= \frac{n}{2} \left( 2 n^2 + 3 n + 1 \right) \\
&= \frac{n}{2} (n + 1)(2n + 1 )
\end{aligned}
よって，
\begin{aligned}
\sum_{k=1}^n k^2 = \frac{1}{6} n (n + 1) (2n + 1) \\
\end{aligned}
数学的帰納法を使って証明してみます．
証明2．
n = 1 のとき，
\begin{aligned}
(左辺) &= \sum_{k=1}^1 k^2 \\
&= 1 \\
(右辺) &= \frac{1}{6} \cdot 1 \cdot (1 + 1) (2 \cdot 1 + 1) \\
&= \frac{1}{6} \cdot 1 \cdot 2 \cdot 3 \\
&= 1
\end{aligned}
等式は成り立つ．
n = m のときに成り立つと仮定すると，
n = m+1 のとき，
\begin{aligned}
\sum_{k=1}^{m+1} k^2 &= \left( \sum_{k=1}^m k^2 \right) + (m + 1)^2 \\
&= \frac{1}{6} m (m + 1) (2m + 1) + (m + 1)^2 \\
&= \frac{1}{6} (m + 1) \{ m(2m + 1) + 6 (m + 1) \} \\
&= \frac{1}{6} (m + 1) ( 2m^2 + 7m + 6) \\
&= \frac{1}{6} (m + 1) ( m + 2 ) ( 2m + 3) \\
&= \frac{1}{6} (m + 1) \{ (m+1) + 1 \} \{ 2(m + 1) + 1 \} \\
\end{aligned}
等式は成り立つ．
よって，全ての自然数 n で等式は成り立つ．
\sum_{k=1}^n k^3
公式．
\begin{aligned}
\sum_{k=1}^n k^3 = \left\{ \frac{1}{2} n (n + 1) \right\}^2 \\
\end{aligned}
証明1．
\begin{aligned}
(k + 1)^4 &= k^4 + 4 k^3 + 6 k^2 + 4 k + 1 \\
(k + 1)^4 - k^4 &= 4 k^3 + 6 k^2 + 4 k + 1
\end{aligned}
となることから，
\begin{aligned}
\sum_{k=1}^n \{(k + 1)^4 - k^4 \} &= \sum_{k=1}^n ( 4 k^3 + 6 k^2 + 4 k + 1 ) \\
\end{aligned}
が成り立つ．
\begin{aligned}
(左辺) &= \sum_{k=1}^n \{(k + 1)^4 - k^4 \} \\
&= \sum_{k=1}^n (k + 1)^4 - \sum_{k=1}^n k^4 \\
&= (n + 1)^4 - 1^4 \\
&= n^4 + 4 n^3 + 6 n^2 + 4 n
\end{aligned}
\begin{aligned}
(右辺) &= \sum_{k=1}^n ( 4 k^3 + 6 k^2 + 4 k + 1 ) \\
&= 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 + 6 \cdot \frac{1}{6} n (n + 1) (2n + 1) + 4 \cdot \frac{1}{2} n (n + 1) + n \\
&= 4 \sum_{k=1}^n k^3 + 2 n^3 + 5 n^2 +4n
\end{aligned}
したがって，
\begin{aligned}
\sum_{k=1}^n \{(k + 1)^4 - k^4 \} &= \sum_{k=1}^n ( 4 k^3 + 6 k^2 + 4 k + 1 ) \\
n^4 + 4 n^3 + 6 n^2 + 4 n &= 4 \sum_{k=1}^n k^3 + 2 n^3 + 5 n^2 +4n \\
\end{aligned}
となる．
\begin{aligned}
4 \sum_{k=1}^n k^3 &= n^4 + 4 n^3 + 6 n^2 + 4 n - 2 n^3 - 5 n^2 - 4n \\
&= n^4 + 2 n^3 + n^2 \\
&= n^2 \left( n + 1 \right)^2 \\
&= \left\{ (n (n + 1) \right\}^2 \\
\end{aligned}
よって，
\begin{aligned}
\sum_{k=1}^n k^3 = \left\{ \frac{1}{2} n (n + 1) \right\}^2 \\
\end{aligned}
数学的帰納法を使った証明もしておきます．
証明2．
n = 1 のとき，
\begin{aligned}
(左辺) &= \sum_{k=1}^1 k^3 \\
&= 1 \\
(右辺) &= \left\{ \frac{1}{2} \cdot 1 \cdot (1 + 1) \right\}^2 \\
&= 1^2 \\
&= 1
\end{aligned}
等式は成り立つ．
n = m のときに成り立つと仮定すると，
n = m+1 のとき，
\begin{aligned}
\sum_{k=1}^{m+1} k^3 &= \left( \sum_{k=1}^m k^3 \right) + (m + 1)^3 \\
&= \left( \frac{1}{2} m^2 + \frac{1}{2} m \right)^2 + (m + 1)^3 \\
&= \frac{1}{4} m^2 (m + 1)^2 + (m + 1)^3 \\
&= (m + 1)^2 \left\{ \frac{1}{4} m^2 + (m + 1) \right\} \\
&= (m + 1)^2 \left(\frac{1}{2} m + 1 \right)^2 \\
&= \left\{ (m + 1) \left(\frac{1}{2} m + 1 \right) \right\}^2 \\
&= \left( \frac{1}{2} m^2 + \frac{3}{2} m + 1 \right)^2 \\
&= \left\{ \frac{1}{2} \left( m^2 + 3 m + 2 \right) \right\}^2 \\
&= \left\{ \frac{1}{2} (m + 1) (m + 2) \right\}^2 \\
&= \left\{ \frac{1}{2} (m + 1) \{(m + 1) + 1 \} \right\}^2 \\
\end{aligned}
等式は成り立つ．
よって，全ての自然数 n で等式は成り立つ．
