数学 - 「離散的」な世界 - 数列 – 数列とその和 - 平方の和、立方の和(一般項)

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数学読本〈3〉平面上のベクトル/複素数と数列とその和/空間図形/2次曲線/数列(松坂和夫(著)、岩波書店)の第13章(「離散的」な世界 - 数列)、13.1(数列とその和)、平方の和、立方の和、問20、21.を取り組んでみる。

問20.

$3 \frac{n (n + 1)}{2} - 2 n = \frac{n (3 n - 1)}{2}$
$\begin{array}{l} \sum_{i = 1}^{n} (2 i^{2} + i - 1) \\ = \frac{2 n (2 n + 1) (n + 1)}{6} + \frac{n (n + 1)}{2} - n \\ = \frac{n (4 n^{2} + 6 n + 2 + 3 n + 3 - 6)}{6} \\ = \frac{n (4 n^{2} + 9 n - 1)}{6} \end{array}$
$\begin{array}{l} \sum_{i = 1}^{n} (i^{2} + 2 i) \\ = \frac{n (2 n + 1) (n + 1)}{6} + n (n + 1) \\ = \frac{n (n + 1) (2 n + 7)}{6} \end{array}$
$\begin{array}{l} \frac{n (2 n + 1) (n + 1)}{6} + \frac{3 n (n + 1)}{2} - 4 n \\ = \frac{n (2 n^{2} + 3 n + 1 + 9 n + 9 - 24)}{6} \\ = \frac{n (2 n^{2} + 12 n - 14)}{6} \\ = \frac{n (n^{2} + 6 n - 7)}{3} \\ = \frac{n (n + 7) (n - 1)}{3} \end{array}$
$\begin{array}{l} {(\frac{n (n + 1)}{2})}^{2} + \frac{2 n (2 n + 1) (n + 1)}{6} + n \\ = \frac{n^{2} {(n + 1)}^{2}}{4} + \frac{2 n (2 n + 1) (n + 1)}{6} + \frac{n (n + 1)}{2} \\ = \frac{n (n + 1) (3 n (n + 1) + 4 (2 n + 1) + 6)}{12} \\ = \frac{n (n + 1) (3 n^{2} + 3 n + 8 n + 4 + 6)}{12} \\ = \frac{n (n + 1) (3 n^{2} + 11 n + 10)}{12} \\ = \frac{n (n + 1) (n + 2) (3 n + 5)}{12} \end{array}$
$\begin{array}{l} \frac{2 n^{2} {(n + 1)}^{2}}{4} + \frac{3 n (2 n + 1) (n + 1)}{6} + \frac{n (n + 1)}{2} \\ = \frac{n (n + 1) (n (n + 1) + (2 n + 1) + 1)}{2} \\ = \frac{n (n + 1) (n^{2} + 3 n + 2)}{4} \\ = \frac{n {(n + 1)}^{2} (n + 2)}{4} \end{array}$

問21.

$\begin{array}{l} a_{n} = {(2 n - 1)}^{2} = 4 n^{2} - 4 n + 1 \\ \frac{4 n (2 n + 1) (n + 1)}{6} - \frac{4 n (n + 1)}{2} + n \\ = \frac{n (4 n^{2} + 6 n + 2 - 6 n - 6 + 3)}{3} \\ = \frac{n (4 n^{2} - 1)}{3} \\ = \frac{n (2 n + 1) (2 n - 1)}{3} \end{array}$
$\begin{array}{l} a_{n} = {(2 n)}^{2} = 4 n^{2} \\ \frac{4 n (2 n + 1) (n + 1)}{6} = \frac{2 n (2 n + 1) (n + 1)}{3} \end{array}$
$\begin{array}{l} a_{n} = n^{2} (3 n - 1) = 3 n^{3} - n^{2} \\ \frac{3 n^{2} {(n + 1)}^{2}}{4} - \frac{n (2 n + 1) (n + 1)}{6} \\ = \frac{n (n + 1) (9 n (n + 1) - 2 (2 n + 1))}{12} \\ = \frac{n (n + 1) (9 n^{2} + 9 n - 4 n - 2)}{12} \\ = \frac{n (n + 1) (9 n^{2} + 5 n - 2)}{12} \end{array}$
$\begin{array}{l} a_{n} = \frac{n (n + 1)}{2} = \frac{1}{2} (n^{2} + n) \\ \frac{1}{2} (\frac{n (2 n + 1) (n + 1)}{6} + \frac{n (n + 1)}{2}) \\ = \frac{2 n (n + 1) (n + 2)}{12} \\ = \frac{n (n + 1) (n + 2)}{6} \end{array}$
$\begin{array}{l} \frac{\frac{n (n + 1)}{2} (\frac{n (n + 1)}{2} + 1)}{2} \\ = \frac{1}{8} n (n + 1) (n (n + 1) + 2) \\ = \frac{n (n + 1) (n^{2} + n + 2)}{8} \end{array}$

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