数列$\{a_n\}$が、$a_1 = 1$, $a_{2n} = 3a_{2n-1}$, $a_{2n+1} = a_{2n} + 3^{n-1}$ (for $n=1, 2, 3, \dots$) で定義される。 (1) $a_2, a_3, a_4$ を求める。 (2) $a_{2n}$ および $a_{2n-1}$ を $n$ の式で表す。 (3) $\sum_{n=1}^{2m} a_n$ を $m$ の式で表す。

1. 問題の内容

数列$\{a_n\}$が、$a_1 = 1$, $a_{2n} = 3a_{2n-1}$, $a_{2n+1} = a_{2n} + 3^{n-1}$ (for $n=1, 2, 3, \dots$) で定義される。

(1) $a_2, a_3, a_4$ を求める。

(2) $a_{2n}$ および $a_{2n-1}$ を $n$ の式で表す。

(3) $\sum_{n=1}^{2m} a_n$ を $m$ の式で表す。

2. 解き方の手順

(1) $a_2, a_3, a_4$ を求める。

$n=1$ のとき、$a_2 = 3a_1 = 3(1) = 3$.

$a_3 = a_2 + 3^{1-1} = a_2 + 3^0 = 3 + 1 = 4$.

$n=2$ のとき、$a_4 = 3a_3 = 3(4) = 12$.

(2) $a_{2n}$ および $a_{2n-1}$ を $n$ の式で表す。

$a_{2n} = 3a_{2n-1}$

$a_{2n+1} = a_{2n} + 3^{n-1}$ より、$a_{2n} = a_{2n-1} + 3^{n-1}$　と考えることもできる．

$a_1 = 1$

$a_2 = 3a_1 = 3$

$a_3 = a_2 + 3^0 = 3 + 1 = 4$

$a_4 = 3a_3 = 3(4) = 12$

$a_5 = a_4 + 3^1 = 12 + 3 = 15$

$a_6 = 3a_5 = 3(15) = 45$

$a_{2n} = 3a_{2n-1}$より、$a_{2n-1} = \frac{1}{3} a_{2n}$.

$a_{2n+1} = a_{2n} + 3^{n-1}$より、$a_{2n} = a_{2n-1} + 3^{n-1}$.

$a_{2n} = 3a_{2n-1}$

$a_{2n-2} = 3a_{2n-3}$

$a_{2n} = a_{2n-1} + 3^{n-1}$

$a_{2n-1} = a_{2n-2} + 3^{n-2}$

$a_{2n} = a_{2n-1} + 3^{n-1}$

$3a_{2n-1} = a_{2n-2} + 3^{n-2} + 3^{n-1}$

$a_{2n} = a_{2n-1} + 3^{n-1}$

$a_{2n-1} = a_{2n-2} + 3^{n-2} = 3a_{2n-3} + 3^{n-2}$

$a_{2n} = 3a_{2n-1} = 3(a_{2n-2} + 3^{n-2})$

$a_{2n} = 3a_{2n-1}$

$a_{2n+1} = a_{2n} + 3^{n-1}$

$a_{2n+1} = 3a_{2n-1} + 3^{n-1}$

$a_{2n+1} = a_{2n} + 3^{n-1}$

$a_1 = 1$

$a_3 = a_2 + 3^0 = 3a_1 + 1 = 3 + 1 = 4$

$a_5 = a_4 + 3^1 = 3a_3 + 3 = 3(4) + 3 = 15$

$a_7 = a_6 + 3^2 = 3a_5 + 9 = 3(15) + 9 = 54$

$a_{2n-1} = 3^{n-1} + 3^{n-2} + \dots + 3^0 = \sum_{k=0}^{n-1} 3^k = \frac{3^n - 1}{3-1} = \frac{3^n - 1}{2}$

$a_{2n} = 3a_{2n-1} = 3 \cdot \frac{3^n - 1}{2} = \frac{3^{n+1} - 3}{2}$

(3) $\sum_{n=1}^{2m} a_n$ を $m$ の式で表す。

$\sum_{n=1}^{2m} a_n = \sum_{n=1}^m a_{2n-1} + \sum_{n=1}^m a_{2n}$

$\sum_{n=1}^m a_{2n-1} = \sum_{n=1}^m \frac{3^n - 1}{2} = \frac{1}{2} \sum_{n=1}^m (3^n - 1) = \frac{1}{2} (\sum_{n=1}^m 3^n - \sum_{n=1}^m 1)$

$= \frac{1}{2} (\frac{3(3^m - 1)}{3-1} - m) = \frac{1}{2} (\frac{3}{2} (3^m - 1) - m) = \frac{3}{4}(3^m - 1) - \frac{m}{2} = \frac{3 \cdot 3^m - 3 - 2m}{4}$

$\sum_{n=1}^m a_{2n} = \sum_{n=1}^m \frac{3^{n+1} - 3}{2} = \frac{1}{2} \sum_{n=1}^m (3^{n+1} - 3) = \frac{1}{2} (\sum_{n=1}^m 3^{n+1} - \sum_{n=1}^m 3)$

$= \frac{1}{2} (\frac{9(3^m - 1)}{3-1} - 3m) = \frac{1}{2} (\frac{9}{2} (3^m - 1) - 3m) = \frac{9}{4}(3^m - 1) - \frac{3m}{2} = \frac{9 \cdot 3^m - 9 - 6m}{4}$

$\sum_{n=1}^{2m} a_n = \frac{3 \cdot 3^m - 3 - 2m}{4} + \frac{9 \cdot 3^m - 9 - 6m}{4} = \frac{12 \cdot 3^m - 12 - 8m}{4} = 3 \cdot 3^m - 3 - 2m = 3^{m+1} - 2m - 3$

3. 最終的な答え

(1) $a_2 = 3, a_3 = 4, a_4 = 12$

(2) $a_{2n-1} = \frac{3^n - 1}{2}$, $a_{2n} = \frac{3^{n+1} - 3}{2}$

(3) $\sum_{n=1}^{2m} a_n = 3^{m+1} - 2m - 3$