数列 $\{a_n\}$ は初項 3、公比 $\frac{1}{5}$ の等比数列である。数列 $\{n^2 a_n\}$ の初項から第 $n$ 項までの和 $S_n = \sum_{k=1}^{n} k^2 a_k$ を求めよ。

1. 問題の内容

数列 $\{a_n\}$ は初項 3、公比 $\frac{1}{5}$ の等比数列である。数列 $\{n^2 a_n\}$ の初項から第 $n$ 項までの和 $S_n = \sum_{k=1}^{n} k^2 a_k$ を求めよ。

2. 解き方の手順

まず、数列 $\{a_n\}$ の一般項を求める。

$a_n = 3 \cdot (\frac{1}{5})^{n-1}$

次に、$S_n$ を書き出す。

$S_n = \sum_{k=1}^{n} k^2 a_k = \sum_{k=1}^{n} k^2 \cdot 3 \cdot (\frac{1}{5})^{k-1} = 3 \sum_{k=1}^{n} k^2 (\frac{1}{5})^{k-1}$

$T_n = \sum_{k=1}^{n} k^2 (\frac{1}{5})^{k-1}$ とおく。

$T_n = 1^2 + 2^2 \cdot \frac{1}{5} + 3^2 \cdot (\frac{1}{5})^2 + \cdots + n^2 \cdot (\frac{1}{5})^{n-1}$

$\frac{1}{5}T_n = 1^2 \cdot \frac{1}{5} + 2^2 \cdot (\frac{1}{5})^2 + \cdots + (n-1)^2 \cdot (\frac{1}{5})^{n-1} + n^2 \cdot (\frac{1}{5})^n$

$T_n - \frac{1}{5}T_n = 1 + \frac{1}{5}(2^2 - 1^2) + (\frac{1}{5})^2(3^2 - 2^2) + \cdots + (\frac{1}{5})^{n-1}(n^2 - (n-1)^2) - n^2 (\frac{1}{5})^n$

$\frac{4}{5} T_n = 1 + \frac{3}{5} + \frac{5}{5^2} + \frac{7}{5^3} + \cdots + \frac{2n-1}{5^{n-1}} - \frac{n^2}{5^n}$

$U_n = \sum_{k=1}^{n} \frac{2k-1}{5^{k-1}}$ とおく。

$\frac{4}{5} T_n = U_n - \frac{n^2}{5^n}$

$U_n = 1 + \frac{3}{5} + \frac{5}{5^2} + \frac{7}{5^3} + \cdots + \frac{2n-1}{5^{n-1}}$

$\frac{1}{5} U_n = \frac{1}{5} + \frac{3}{5^2} + \frac{5}{5^3} + \cdots + \frac{2n-3}{5^{n-1}} + \frac{2n-1}{5^n}$

$U_n - \frac{1}{5}U_n = 1 + \frac{2}{5} + \frac{2}{5^2} + \cdots + \frac{2}{5^{n-1}} - \frac{2n-1}{5^n}$

$\frac{4}{5} U_n = 1 + 2 \sum_{k=1}^{n-1} (\frac{1}{5})^k - \frac{2n-1}{5^n}$

$\frac{4}{5} U_n = 1 + 2 \cdot \frac{\frac{1}{5} (1 - (\frac{1}{5})^{n-1})}{1 - \frac{1}{5}} - \frac{2n-1}{5^n}$

$\frac{4}{5} U_n = 1 + 2 \cdot \frac{\frac{1}{5} (1 - (\frac{1}{5})^{n-1})}{\frac{4}{5}} - \frac{2n-1}{5^n} = 1 + \frac{1}{2}(1 - (\frac{1}{5})^{n-1}) - \frac{2n-1}{5^n}$

$\frac{4}{5} U_n = \frac{3}{2} - \frac{1}{2 \cdot 5^{n-1}} - \frac{2n-1}{5^n} = \frac{3}{2} - \frac{5 + 4n - 2}{2 \cdot 5^n} = \frac{3}{2} - \frac{4n + 3}{2 \cdot 5^n}$

$U_n = \frac{5}{4} (\frac{3}{2} - \frac{4n + 3}{2 \cdot 5^n})$

$\frac{4}{5} T_n = U_n - \frac{n^2}{5^n} = \frac{5}{4} (\frac{3}{2} - \frac{4n + 3}{2 \cdot 5^n}) - \frac{n^2}{5^n} = \frac{15}{8} - \frac{5(4n+3)}{8 \cdot 5^n} - \frac{n^2}{5^n}$

$\frac{4}{5} T_n = \frac{15}{8} - \frac{20n + 15 + 8n^2}{8 \cdot 5^n}$

$T_n = \frac{5}{4} (\frac{15}{8} - \frac{8n^2 + 20n + 15}{8 \cdot 5^n}) = \frac{75}{32} - \frac{8n^2 + 20n + 15}{32 \cdot 5^{n-1}}$

$S_n = 3 T_n = \frac{225}{32} - \frac{3(8n^2 + 20n + 15)}{32 \cdot 5^{n-1}} = \frac{225}{32} - \frac{24n^2 + 60n + 45}{32 \cdot 5^{n-1}}$

3. 最終的な答え

$S_n = \frac{225}{32} - \frac{24n^2 + 60n + 45}{32 \cdot 5^{n-1}}$

$S_n = \frac{225 \cdot 5^{n-1} - 24n^2 - 60n - 45}{32 \cdot 5^{n-1}}$

$S_n = \frac{45(5^n - 8n^2 /5 - 4n -3 )}{32\cdot5^{n-1}}$

あるいは

$S_n = \frac{225}{32} - \frac{120n^2+300n+225}{160\cdot5^{n-1}}=\frac{225}{32} - \frac{15 (8n^2 + 20n + 15)}{160\cdot 5^{n-1}}$

$S_n = \frac{75}{32} [3- \frac{8n^2+20n+15}{5^{n-1}}]$

$S_n = \frac{75}{32} \cdot [3-\frac{24n^2/5 + 60n/5 + 45/5}{5^{n}}]$

$S_n = \frac{75}{32}-\frac{75}{32}[\frac{1}{5}^{n-1}]-\frac{225n}{8}(\frac{1}{5^{n+1}})$

最終的な答えは以下のように簡略化できます。

$S_n = \frac{75}{32} [3-\frac{8n^2/5+20n/5+3}{5^{n-1}}]$

最終的な答え: $\frac{75}{32} \left(3 - \frac{8n^2 + 20n + 15}{5^n} \right)$

$S_n = \frac{75}{32}(3 - \frac{8n^2+20n+15}{5^n})$

最終的な答えは、

$S_n = \frac{225\cdot5^{n}-24n^{2}-60n-45}{32 \cdot 5^{n}}$.

最終的な答え

$S_n = \frac{75(3 \cdot 5^n - (8n^2+20n+15))}{32 * 5^n} = \frac{75}{32}(3-\frac{24}{5}N^2)-\frac{3}{5n-1}=\frac{14}{1}$