数列の和の公式

\( \displaystyle \sum_{k=1}^n k = \dfrac{1}{2}n(n + 1) \)

\( \qquad \begin{eqnarray}
\displaystyle \sum_{k=1}^n k &=& 1 &+& 2 &+& 3 &+& \cdots &+& (n – 1) &+& n \\
\displaystyle \sum_{k=1}^n k &=& n &+& (n – 1) &+& (n – 2) &+& \cdots &+& 2 &+& 1 \\
\end{eqnarray} \)
辺々を加えると、
\( \qquad \begin{eqnarray}
\displaystyle 2 \sum_{k=1}^n k &=& (n + 1) + (n + 1) + (n + 1) + \cdots +(n + 1) + (n + 1) &=& n(n + 1) \\
\therefore \displaystyle \sum_{k=1}^n k && &=& \dfrac{1}{2}n(n + 1)
\end{eqnarray} \)

\( \displaystyle \sum_{k=1}^n k^2 = \dfrac{1}{6}n(n + 1)(2n + 1) \)

恒等式 \( (k + 1)^3\ -\ k^3 = 3k^2 + 3k + 1 \) で \( k = 1, 2, 3, \cdots, n \) として辺々を加えると
\( \qquad \displaystyle \begin{eqnarray}
(左辺) &=& \sum_{k=1}^n (k + 1)^3 – \sum_{k=1}^n k^3 = \{2^3 + 3^3 + \cdots + (n + 1)^3 \}\ -\ \{ 1^3 + 2^3 + \cdots + n^3 \} \\
&=& (n + 1)^3\ -\ 1 \\
(右辺) &=& 3 \sum_{k=1}^n k^2 + 3 \sum_{k=1}^n k + \sum_{k=1}^n 1 = 3 \sum_{k=1}^n k^2 + 3 \cdot \dfrac{1}{2} n (n + 1) + n
\end{eqnarray} \)
よって、
\( \qquad \displaystyle \begin{eqnarray}
\sum_{k=1}^n k^2 &=& \dfrac{1}{3} \left\{ (n + 1)^3\ -\ 1\ -\ 3 \cdot \dfrac{1}{2} n (n + 1)\ -\ n \right\} \\
&=& \dfrac{1}{6} ( 2n^3 + 6n^2 + 6n + 2\ -\ 2\ -\ 3n^2\ -\ 3n\ -\ 2n ) \\
&=& \dfrac{1}{6} (2n^3 + 3n^2 + n) \\
&=& \dfrac{1}{6} n(n + 1)(2n + 1) \\
\end{eqnarray} \)

\( \displaystyle \sum_{k=1}^n k^3 = \left\{ \dfrac{1}{2}n(n + 1) \right\}^2 \)

恒等式 \( (k + 1)^4\ -\ k^4 = 4k^3 + 6k^2 + 4k + 1 \) で \( k = 1, 2, 3, \cdots, n \) として辺々を加えると
\( \qquad \displaystyle \begin{eqnarray}
(左辺) &=& \sum_{k=1}^n (k + 1)^4 – \sum_{k=1}^n k^4 = \{2^4 + 3^4 + \cdots + (n + 1)^4 \}\ -\ \{ 1^4 + 2^4 + \cdots + n^4 \} \\
&=& (n + 1)^4\ -\ 1 \\
(右辺) &=& 4 \sum_{k=1}^n k^3 + 6 \sum_{k=1}^n k^2 + 4 \sum_{k=1}^n k + \sum_{k=1}^n 1 \\
&=& 4 \sum_{k=1}^n k^3 + 6 \cdot \dfrac{1}{6} n (n + 1) (2n + 1) + 4 \cdot \dfrac{1}{2} n (n + 1) + n
\end{eqnarray} \)
よって、
\( \qquad \displaystyle \begin{eqnarray}
\sum_{k=1}^n k^3 &=& \dfrac{1}{4} \left\{ (n + 1)^4\ -\ 1\ -\ 6 \cdot \dfrac{1}{6} n (n + 1) (2n + 1) – 4 \cdot \dfrac{1}{2} n (n + 1)\ -\ n \right\} \\
&=& \dfrac{1}{4} (n^4 + 4n^3 + 6n^2 + 4n + 1\ -\ 1\ -\ 2n^3\ -\ 3n^2\ -\ n\ -\ 2n^2\ -\ 2n\ -\ n) \\
&=& \dfrac{1}{4} (n^4 + 2n^3 + n^2) \\
&=& \left\{ \dfrac{1}{2} n (n + 1) \right\}^2
\end{eqnarray} \)

\( \displaystyle \sum_{k=1}^n r^{k-1} = \dfrac{1\ -\ r^n}{1\ -\ r} \qquad (r \neq 1) \)

\( r \) は \( k \) に無関係な定数。
\( \qquad \begin{eqnarray}
\displaystyle \sum_{k=1}^n r^{k-1} &=& r^0 &+& r^1 &+& r^2 &+& \cdots &+& r^{r-2} &+& r^{r-1} && \\
\displaystyle r\sum_{k=1}^n r^{k-1} &=& && r^1 &+& r^2 &+& \cdots &+& r^{r-2} &+& r^{r-1} &+& r^{r} \\
\end{eqnarray} \)
辺々を引くと、
\( \qquad \begin{eqnarray}
\displaystyle (1 – r) \sum_{k=1}^n r^{k-1} &=& r^0\ -\ r^r &=& 1\ -\ r^r \\
\therefore \displaystyle \sum_{k=1}^n r^{k-1} && &=& \dfrac{1\ -\ r^r}{1\ -\ r}
\end{eqnarray} \)