数列 $\{a_n\}$ が、$a_1 = 1$, $a_{n+1} = (\sqrt{2}-1)a_n + 2$ ($n = 1, 2, 3, \dots$) で定義される。このとき、$a_n = p_n + \sqrt{2} q_n$ と表せる。以下の問いに答える。 (1) $p_1, p_2, q_1, q_2$ を求めよ。 (2) $p_{n+1}, q_{n+1}$ を $p_n, q_n$ を用いてそれぞれ表せ。 (3) $b_n = a_n - (2 + \sqrt{2})$ とする。$b_{n+1}$ を $b_n$ を用いて表せ。また、$b_n$ を $n$ を用いて表せ。 (4) $c_n = p_n - \sqrt{2} q_n$ とする。$c_{n+1}$ を $c_n$ を用いて表せ。 (5) $p_n, q_n$ を $n$ を用いてそれぞれ表せ。

1. 問題の内容

数列 $\{a_n\}$ が、$a_1 = 1$, $a_{n+1} = (\sqrt{2}-1)a_n + 2$ ($n = 1, 2, 3, \dots$) で定義される。このとき、$a_n = p_n + \sqrt{2} q_n$ と表せる。以下の問いに答える。

(1) $p_1, p_2, q_1, q_2$ を求めよ。

(2) $p_{n+1}, q_{n+1}$ を $p_n, q_n$ を用いてそれぞれ表せ。

(3) $b_n = a_n - (2 + \sqrt{2})$ とする。$b_{n+1}$ を $b_n$ を用いて表せ。また、$b_n$ を $n$ を用いて表せ。

(4) $c_n = p_n - \sqrt{2} q_n$ とする。$c_{n+1}$ を $c_n$ を用いて表せ。

(5) $p_n, q_n$ を $n$ を用いてそれぞれ表せ。

2. 解き方の手順

(1)

$a_1 = 1 = p_1 + \sqrt{2}q_1$ より、$p_1 = 1$, $q_1 = 0$.

$a_2 = (\sqrt{2}-1)a_1 + 2 = (\sqrt{2}-1)(1) + 2 = 1 + \sqrt{2}$.

$a_2 = p_2 + \sqrt{2}q_2$ より、$p_2 = 1$, $q_2 = 1$.

(2)

$a_{n+1} = (\sqrt{2}-1)a_n + 2 = (\sqrt{2}-1)(p_n + \sqrt{2}q_n) + 2 = (\sqrt{2}-1)p_n + 2q_n - \sqrt{2}q_n + 2 = (2-p_n+2q_n) + \sqrt{2}(p_n-q_n)$

また、$a_{n+1} = p_{n+1} + \sqrt{2}q_{n+1}$

よって、$p_{n+1} = 2-p_n+2q_n$ , $q_{n+1} = p_n - q_n$

(3)

$b_n = a_n - (2 + \sqrt{2})$ より、$a_n = b_n + 2 + \sqrt{2}$.

$a_{n+1} = b_{n+1} + 2 + \sqrt{2}$.

$a_{n+1} = (\sqrt{2}-1)a_n + 2$ に代入すると、

$b_{n+1} + 2 + \sqrt{2} = (\sqrt{2}-1)(b_n + 2 + \sqrt{2}) + 2 = (\sqrt{2}-1)b_n + 2\sqrt{2} - 2 + 2 + 2\sqrt{2} - 2 + 2 = (\sqrt{2}-1)b_n + 4\sqrt{2}-2$

$b_{n+1} = (\sqrt{2}-1)b_n + 3\sqrt{2}-4$

$b_1 = a_1 - (2+\sqrt{2}) = 1 - (2+\sqrt{2}) = -1 - \sqrt{2}$

$b_{n+1} = (\sqrt{2}-1)b_n$

$b_n = b_1 (\sqrt{2}-1)^{n-1} = (-1-\sqrt{2})(\sqrt{2}-1)^{n-1} = -(\sqrt{2}+1)(\sqrt{2}-1)^{n-1} = -(\sqrt{2}-1)^{-1} (\sqrt{2}-1)^{n-1} = -(\sqrt{2}-1)^{n-2}$

(4)

$c_n = p_n - \sqrt{2} q_n$

$c_{n+1} = p_{n+1} - \sqrt{2} q_{n+1} = (2-p_n+2q_n) - \sqrt{2} (p_n - q_n) = 2 - p_n + 2q_n - \sqrt{2} p_n + \sqrt{2} q_n = 2 - (p_n - \sqrt{2} q_n) - \sqrt{2}(p_n - \sqrt{2}q_n) = 2 - (1+\sqrt{2}) (p_n - \sqrt{2}q_n) = 2 - (1+\sqrt{2}) c_n$

(5)

$a_n = p_n + \sqrt{2}q_n$, $b_n = a_n - (2+\sqrt{2})$ より、$a_n = b_n + (2+\sqrt{2})$

$b_n = -(\sqrt{2}-1)^{n-2}$ より、$a_n = 2 + \sqrt{2} - (\sqrt{2}-1)^{n-2}$

$c_n = p_n - \sqrt{2}q_n$

$p_n = \frac{a_n + c_n}{2}$, $q_n = \frac{a_n - c_n}{2\sqrt{2}}$

$a_n - c_n = (p_n + \sqrt{2}q_n) - (p_n - \sqrt{2}q_n) = 2\sqrt{2}q_n$

$a_n + c_n = (p_n + \sqrt{2}q_n) + (p_n - \sqrt{2}q_n) = 2p_n$

$c_1 = p_1 - \sqrt{2} q_1 = 1 - \sqrt{2} (0) = 1$

$c_{n+1} = 2 - (1+\sqrt{2})c_n$ より

$c_n = \frac{2}{2+\sqrt{2}} + \alpha (-(1+\sqrt{2}))^{n-1}$ , where $2- (1+\sqrt{2}) \frac{2}{2+\sqrt{2}} = \frac{2(2+\sqrt{2})-2(1+\sqrt{2})}{2+\sqrt{2}} = \frac{2+0}{2+\sqrt{2}} = \frac{2}{2+\sqrt{2}}$

$c_n = \sqrt{2}$

3. 最終的な答え

(1) $p_1 = 1$, $p_2 = 1$, $q_1 = 0$, $q_2 = 1$

(2) $p_{n+1} = 2-p_n+2q_n$, $q_{n+1} = p_n - q_n$

(3) $b_{n+1} = (\sqrt{2}-1)b_n$, $b_n = -(\sqrt{2}-1)^{n-2}$

(4) $c_{n+1} = 2 - (1+\sqrt{2}) c_n$

(5) $p_n = \frac{2 + \sqrt{2} - (\sqrt{2}-1)^{n-2}+\sqrt{2}}{2}$, $q_n = \frac{2+\sqrt{2} - (\sqrt{2}-1)^{n-2}-\sqrt{2}}{2\sqrt{2}}$

$p_n = \frac{2 + \sqrt{2}}{2} - \frac{(\sqrt{2}-1)^{n-2}}{2}$, $q_n = \frac{2+\sqrt{2}}{2\sqrt{2}} - \frac{(\sqrt{2}-1)^{n-2}}{2\sqrt{2}}$

$p_n = 1 + \frac{\sqrt{2}}{2} - \frac{(\sqrt{2}-1)^{n-2}}{2}$, $q_n = \frac{1}{\sqrt{2}} + \frac{1}{2} - \frac{(\sqrt{2}-1)^{n-2}}{2\sqrt{2}}$