数列と関数数列数列の和累乗

学習環境

現代数学への入門微分と積分1 初等関数を中心に (青本和彦(著)、岩波書店)の第1章(数列と関数)、1.1(数列)、c(数列の和)の問13の解答を求めてみる。

m = 1

おけば、

\begin{array}{l} a_{n} = n \\ S_{N} = \frac{N (N + 1)}{1 + 1} = \frac{N (N + 1)}{2} \end{array}

よって、

\begin{array}{l} k = 1 \\ S_{N}^{1} = \sum_{n = 1}^{N} n = \sum_{n = 1}^{N} a_{n} = \frac{N (N + 1)}{2} \end{array}

\begin{array}{l} k = 2 \\ m = 2 \\ a_{n} = n (n + 1) = n^{2} + n \\ S_{N} = \frac{N (N + 1) (N + 2)}{3} \\ S_{N}^{2} = \sum_{n = 1}^{N} n^{2} \\ = \sum_{n = 1}^{N} (a_{n} - n) \\ = \sum_{n = 1}^{N} a_{n} - \sum_{n = 1}^{N} n \\ = S_{N} - S_{N}^{1} \\ = \frac{N (N + 1) (N + 2)}{3} - \frac{N (N + 1)}{2} \\ = \frac{N (N + 1) (2 N + 4 - 3)}{6} \\ = \frac{N (N + 1) (2 N + 1)}{6} \end{array}

\begin{array}{l} k = 3 \\ m = 3 \\ a_{n} = n (n + 1) (n + 2) \\ = (n^{2} + n) (n + 2) \\ = n^{3} + 3 n^{2} + 2 n \\ S_{N} = \frac{N (N + 1) (N + 2) (N + 3)}{4} \\ S_{N}^{3} = \sum_{n = 1}^{N} n^{3} \\ = \sum_{n = 1}^{N} a_{n} - 3 \sum_{n = 1}^{N} n^{2} - 2 \sum_{n = 1}^{N} n \\ = S_{N} - 3 S_{N}^{2} - 2 S_{N}^{1} \\ = \frac{N (N + 1) (N + 2) (N + 3)}{4} - \frac{N (N + 1) (2 N + 1)}{2} - N (N + 1) \\ = \frac{N (N + 1) ((N + 2) (N + 3) - 2 (2 N + 1) - 4)}{4} \\ = \frac{N (N + 1) (N^{2} + N)}{4} \\ = \frac{N^{2} {(N + 1)}^{2}}{4} \\ = {(\frac{N (N + 1)}{2})}^{2} \\ = {(S_{N}^{1})}^{2} \end{array}

\begin{array}{l} k = 4 \\ m = 4 \\ a_{n} = n (n + 1) (n + 2) (n + 3) \\ = (n^{2} + n) (n^{2} + 5 n + 6) \\ = n^{4} + 6 n^{3} + 11 n^{2} + 6 n \\ S_{N} = \frac{1}{5} \prod_{k = 0}^{4} (N + k) \\ S_{N}^{4} = \sum_{n = 1}^{N} n^{4} \\ = \sum_{n = 1}^{N} a_{n} - 6 \sum_{n = 1}^{N} n^{3} - 11 \sum_{n = 1}^{N} n^{2} - 6 \sum_{n = 1}^{N} n \\ = \frac{1}{5} \prod_{k = 0}^{4} (N + k) - \frac{3 N^{2} {(N + 1)}^{2}}{2} - \frac{11 N (N + 1) (2 N + 1)}{6} - 3 N (N + 1) \\ = \frac{N (N + 1) (6 (N + 2) (N + 3) (N + 4) - 45 N (N + 1) - 55 (2 N + 1) - 90)}{30} \\ = \frac{N (N + 1) (6 N^{3} + 9 N^{2} + N - 1)}{30} \end{array}