数列 $\{a_n\}$ が以下の漸化式で定義される。 $a_1 = 1$, $a_{2n} = 2a_{2n-1}$, $a_{2n+1} = a_{2n} + 2^{n-1}$ (1) $a_{2n-1}, a_{2n}$ をそれぞれ求めよ。 (2) $\sum_{k=1}^{n} a_k$ を求めよ。

1. 問題の内容

数列 $\{a_n\}$ が以下の漸化式で定義される。

$a_1 = 1$,

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

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

(1) $a_{2n-1}, a_{2n}$ をそれぞれ求めよ。

(2) $\sum_{k=1}^{n} a_k$ を求めよ。

2. 解き方の手順

(1) まず、$a_{2n}$ と $a_{2n+1}$ の関係式から $a_{2n}$ を消去する。

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

したがって、

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

ここで、$b_n = a_{2n-1}$ とおくと、

$b_{n+1} = 2b_n + 2^{n-1}$

この漸化式を解く。両辺を $2^{n+1}$ で割ると、

$\frac{b_{n+1}}{2^{n+1}} = \frac{b_n}{2^n} + \frac{2^{n-1}}{2^{n+1}} = \frac{b_n}{2^n} + \frac{1}{4}$

$c_n = \frac{b_n}{2^n}$ とおくと、

$c_{n+1} = c_n + \frac{1}{4}$

これは公差 $\frac{1}{4}$ の等差数列であり、

$c_1 = \frac{b_1}{2^1} = \frac{a_1}{2} = \frac{1}{2}$

よって、

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

したがって、

$b_n = 2^n c_n = \frac{n+1}{4} 2^n = (n+1)2^{n-2}$

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

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

(2) $\sum_{k=1}^n a_k$ を求める。

$\sum_{k=1}^{2n} a_k = \sum_{k=1}^n a_{2k-1} + \sum_{k=1}^n a_{2k} = \sum_{k=1}^n (k+1)2^{k-2} + \sum_{k=1}^n (k+1)2^{k-1}$

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

$= 3 \sum_{k=1}^n k2^{k-2} + 3\sum_{k=1}^n 2^{k-2} = 3 \sum_{k=1}^n k2^{k-2} + 3 \frac{\frac{1}{2}(2^{n-1}-1)}{2-1} = 3 \sum_{k=1}^n k2^{k-2} + \frac{3}{2} (2^{n-1}-1)$

$S = \sum_{k=1}^n k2^{k-2} = \frac{1}{4} + \frac{2}{2} + \frac{3}{4}2 + \frac{4}{4}2^2 + \cdots + \frac{n}{4}2^{n-2}$

$2S = \frac{1}{2} + \frac{2}{4}2 + \frac{3}{4}2^2 + \cdots + \frac{n}{4}2^{n-1}$

$S - 2S = \frac{1}{4} + \frac{1}{2} + \frac{1}{4}2 + \frac{1}{4}2^2 + \cdots + \frac{1}{4}2^{n-2} - \frac{n}{4}2^{n-1}$

$-S = \frac{1}{4} + \frac{\frac{1}{2}(2^{n-2}-1)}{2-1} - \frac{n}{4}2^{n-1} = \frac{1}{4} + \frac{1}{2}(2^{n-2}-1) - \frac{n}{4}2^{n-1}$

$-S = \frac{1}{4} + \frac{1}{2}2^{n-2} - \frac{1}{2} - \frac{n}{4}2^{n-1} = \frac{1}{4} 2^{n-1} - \frac{1}{4} - \frac{n}{4}2^{n-1}$

$S = \frac{n-1}{4}2^{n-1} + \frac{1}{4}$

$\sum_{k=1}^{2n} a_k = 3(\frac{n-1}{4}2^{n-1} + \frac{1}{4}) + \frac{3}{2} (2^{n-1}-1) = \frac{3(n-1)}{4} 2^{n-1} + \frac{3}{4} + \frac{3}{2}2^{n-1} - \frac{3}{2}$

$= \frac{3n-3+6}{4} 2^{n-1} - \frac{3}{4} = \frac{3n+3}{4} 2^{n-1} - \frac{3}{4} = \frac{3(n+1)}{4} 2^{n-1} - \frac{3}{4} = \frac{3(n+1)2^{n-1}-3}{4}$

3. 最終的な答え

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

(2) $\sum_{k=1}^{2n} a_k = \frac{3(n+1)2^{n-1}-3}{4}$

$\sum_{k=1}^n a_k$ は $n$が偶数の場合のみ答えられる。

$\sum_{k=1}^n a_k = \frac{3(\frac{n}{2}+1)2^{\frac{n}{2}-1}-3}{4}$ if $n$ is even.

If $n$ is odd:

$\sum_{k=1}^{n} a_k = \sum_{k=1}^{n-1} a_k + a_n = \frac{3(\frac{n-1}{2}+1)2^{\frac{n-1}{2}-1}-3}{4} + a_n$.

If $n=2m+1$:

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

$\sum_{k=1}^n a_k = \frac{3(\frac{n-1}{2}+1)2^{\frac{n-1}{2}-1}-3}{4} + a_n = \frac{3(\frac{n+1}{2})2^{\frac{n-3}{2}}-3}{4} + (\frac{n+3}{2})2^{\frac{n-3}{2}} = \frac{3(n+1)2^{\frac{n-3}{2}}-6 + 4(n+3)2^{\frac{n-3}{2}}}{8} = \frac{(7n+15)2^{\frac{n-3}{2}}-6}{8}$

最終的な答え：

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

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

$\sum_{k=1}^{2n} a_k = \frac{3(n+1)2^{n-1}-3}{4}$

$\sum_{k=1}^{2n+1} a_k = \frac{(7(2n+1)+15)2^{\frac{2n+1-3}{2}} - 6}{8} = \frac{(14n+22)2^{n-1}-6}{8} = \frac{(7n+11)2^{n-1}-3}{4}$

$\sum_{k=1}^n a_k = \begin{cases}

\frac{3(\frac{n}{2}+1)2^{\frac{n}{2}-1}-3}{4} & \text{if n is even} \\

\frac{(7(\frac{n-1}{2})+11)2^{\frac{n-1}{2}-1}-3}{4} & \text{if n is odd}

\end{cases} $

$\sum_{k=1}^{n} a_k = \begin{cases}

\frac{3(\frac{n}{2}+1)2^{\frac{n}{2}-1}-3}{4} & \text{if } n \text{ is even} \\

\frac{(7(\frac{n-1}{2})+11)2^{\frac{n-3}{2}}-3}{4} & \text{if } n \text{ is odd}

\end{cases} $

$\sum_{k=1}^n a_k = \begin{cases}

\frac{3(\frac{n}{2}+1)2^{\frac{n}{2}-1}-3}{4}, & n \text{ is even} \\

\frac{(7(\frac{n-1}{2})+11)2^{\frac{n-1}{2}-1}-3}{4} & n \text{ is odd}

\end{cases}$

$\sum_{k=1}^{n} a_k = \begin{cases}

\frac{3(\frac{n}{2}+1)2^{\frac{n}{2}-1}-3}{4} & \text{if } n \text{ is even} \\

\frac{1}{4}((7(\frac{n-1}{2})+11)2^{\frac{n-1}{2}-1}-3) & \text{if } n \text{ is odd}

\end{cases}$

$\sum_{k=1}^n a_k=\begin{cases} \frac{3(\frac{n}{2}+1)2^{\frac{n}{2}-1}-3}{4} & (n:even)\\ \frac{(7(n-1)+22)2^{\frac{n-3}{2}}-6}{8} & (n:odd) \end{cases}$

$\sum_{k=1}^n a_k=\begin{cases} \frac{3(\frac{n}{2}+1)2^{\frac{n}{2}-1}-3}{4} & (n:even)\\ \frac{(7n+15)2^{\frac{n-3}{2}}-6}{8} & (n:odd) \end{cases}$

$\sum_{k=1}^n a_k=\begin{cases} \frac{3(\frac{n}{2}+1)2^{\frac{n}{2}-1}-3}{4} & \text{if }n\text{ is even}\\ \frac{(7n+15)2^{\frac{n-3}{2}}-6}{8} & \text{if }n\text{ is odd} \end{cases}$

```

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

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

\sum_{k=1}^n a_k = \begin{cases}

\frac{3(\frac{n}{2}+1)2^{\frac{n}{2}-1}-3}{4} & \text{if } n \text{ is even} \\

\frac{(7n+15)2^{\frac{n-3}{2}}-6}{8} & \text{if } n \text{ is odd}

\end{cases}

```