数学 - 「離散的」な世界 - 数列 – 数学的帰納法と数列 - 帰納的定義(一般項)

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数学読本〈3〉平面上のベクトル/複素数と数学的帰納法と数列/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)、13.2(数学的帰納法と数列)、数学的帰納法、問32、33.を取り組んでみる。

問32.

$\begin{array}{l} b_{n} = a_{n + 1} - a_{n} \\ = a_{n} + 2 n - a_{n} \\ = 2 n \\ n > 1 \\ a_{n} = a_{1} + \sum_{i = 1}^{n - 1} b_{i} \\ = 1 + 2 \cdot \frac{(n - 1) n}{2} \\ = n^{2} - n + 1 \\ 1^{2} - 1 + 1 = 1 \\ a_{n} = n^{2} - n + 1 \end{array}$
$\begin{array}{l} b_{n} = a_{n + 1} - a_{n} \\ = a_{n} + 3^{n} - a_{n} \\ = 3^{n} \\ n > 1 \\ a_{n} = a_{1} + \sum_{i = 1}^{n - 1} b_{n} \\ = 2 + \frac{3 (3^{n - 1} - 1)}{3 - 1} \\ = \frac{3^{n}}{2} + \frac{1}{2} \\ \frac{3}{2} + \frac{1}{2} = 2 \\ a_{n} = \frac{1}{2} (3^{n} + 1) \end{array}$
$\begin{array}{l} x = 3 x - 2 \\ x = 1 \\ a_{n + 1} - 1 = 3 (a_{n} - 1) \\ a_{n} - 1 = (2 - 1) \cdot 3^{n - 1} \\ a_{n} = 3^{n - 1} + 1 \end{array}$
$\begin{array}{l} x = - 2 x + 9 \\ x = 3 \\ a_{n + 1} - 3 = - 2 (x - 3) \\ a_{n} - 3 = (1 - 3) \cdot {(- 2)}^{n - 1} \\ a_{n} = {(- 2)}^{n} + 3 \end{array}$
$\begin{array}{l} 2 x = x - 6 \\ x = - 6 \\ 2 (a_{n + 1} + 6) = a_{n} + 6 \\ a_{n + 1} + 6 = \frac{1}{2} (a_{n} + 6) \\ a_{n} + 6 = (2 + 6) \cdot \frac{1}{2^{n - 1}} \\ a_{n} = \frac{1}{2^{n - 4}} - 6 \end{array}$
$\begin{array}{l} 1, \frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \frac{1}{5} \\ a_{n + 1} = \frac{n}{n + 1} \cdot \frac{1}{n} = \frac{1}{n + 1} \\ a_{n} = \frac{1}{n} \end{array}$
$\begin{array}{l} 2, \frac{2}{3}, \frac{2}{5}, \frac{2}{7}, \frac{2}{9} \\ a_{n + 1} = \frac{\frac{2}{2 n - 1}}{\frac{2}{2 n - 1} + 1} \\ = \frac{2}{2 n - 1} \cdot \frac{2 n - 1}{2 n + 1 `} \\ = \frac{2}{2 (n + 1) - 1} \\ a_{n} = \frac{2}{2 n - 1} \end{array}$
$\begin{array}{l} 3, 3, 6, 12, 24 \\ n > 1 \\ a_{n + 1} = a_{1} + a_{2} + \cdot \cdot \cdot + a_{n - 1} + a_{n} \\ = 2 a_{n} \\ a_{n} = 3 \cdot 2^{n - 2} \\ a_{1} = 3, a_{n} = 3 \cdot 2^{n - 2} (n > 1) \end{array}$

問33.

\begin{array}{l} 1, \frac{1}{3}, \frac{1}{7}, \frac{1}{15}, \frac{1}{31} \\ a_{n + 1} = \frac{\frac{1}{2^{n} - 1}}{\frac{1}{2^{n} - 1} + 2} \\ = \frac{1}{2^{n} - 1} \cdot \frac{2^{n} - 1}{1 + 2^{n + 1} - 2} \\ = \frac{1}{2^{n + 1} - 1} \\ a_{n} = \frac{1}{2^{n} - 1} \end{array}

Kamimura's blog

2017年1月18日水曜日

数学 - 「離散的」な世界 - 数列 – 数学的帰納法と数列 - 帰納的定義(一般項)

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