与えられた式 $\frac{4}{75}S_n = 2\left(\frac{5}{16} - \frac{4n+5}{16}\cdot \frac{1}{5^n}\right) - \frac{1}{4}\left\{1-\left(\frac{1}{5}\right)^n\right\} - n^2\left(\frac{1}{5}\right)^{n+1}$ を計算し、$S_n$ を求める問題です。

1. 問題の内容

与えられた式 $\frac{4}{75}S_n = 2\left(\frac{5}{16} - \frac{4n+5}{16}\cdot \frac{1}{5^n}\right) - \frac{1}{4}\left\{1-\left(\frac{1}{5}\right)^n\right\} - n^2\left(\frac{1}{5}\right)^{n+1}$ を計算し、$S_n$ を求める問題です。

2. 解き方の手順

まず、与えられた式を整理します。

$\frac{4}{75}S_n = 2\left(\frac{5}{16} - \frac{4n+5}{16}\cdot \frac{1}{5^n}\right) - \frac{1}{4}\left\{1-\left(\frac{1}{5}\right)^n\right\} - n^2\left(\frac{1}{5}\right)^{n+1}$

右辺を展開します。

$\frac{4}{75}S_n = \frac{10}{16} - \frac{8n+10}{16}\cdot \frac{1}{5^n} - \frac{1}{4} + \frac{1}{4} \cdot \frac{1}{5^n} - n^2 \cdot \frac{1}{5^{n+1}}$

$\frac{4}{75}S_n = \frac{5}{8} - \frac{1}{4} - \frac{8n+10}{16}\cdot \frac{1}{5^n} + \frac{1}{4} \cdot \frac{1}{5^n} - \frac{n^2}{5} \cdot \frac{1}{5^n}$

$\frac{4}{75}S_n = \frac{5}{8} - \frac{2}{8} - \frac{8n+10}{16}\cdot \frac{1}{5^n} + \frac{4}{16} \cdot \frac{1}{5^n} - \frac{n^2}{5} \cdot \frac{1}{5^n}$

$\frac{4}{75}S_n = \frac{3}{8} - \frac{8n+10-4}{16}\cdot \frac{1}{5^n} - \frac{n^2}{5} \cdot \frac{1}{5^n}$

$\frac{4}{75}S_n = \frac{3}{8} - \frac{8n+6}{16}\cdot \frac{1}{5^n} - \frac{n^2}{5} \cdot \frac{1}{5^n}$

$\frac{4}{75}S_n = \frac{3}{8} - \left(\frac{8n+6}{16} + \frac{n^2}{5}\right) \frac{1}{5^n}$

$\frac{4}{75}S_n = \frac{3}{8} - \left(\frac{5(8n+6) + 16n^2}{80}\right) \frac{1}{5^n}$

$\frac{4}{75}S_n = \frac{3}{8} - \left(\frac{40n+30 + 16n^2}{80}\right) \frac{1}{5^n}$

$\frac{4}{75}S_n = \frac{3}{8} - \left(\frac{16n^2+40n+30}{80}\right) \frac{1}{5^n}$

$\frac{4}{75}S_n = \frac{3}{8} - \left(\frac{8n^2+20n+15}{40}\right) \frac{1}{5^n}$

両辺に $\frac{75}{4}$ をかけます。

$S_n = \frac{75}{4}\left[\frac{3}{8} - \left(\frac{8n^2+20n+15}{40}\right) \frac{1}{5^n}\right]$

$S_n = \frac{225}{32} - \frac{75}{4} \cdot \left(\frac{8n^2+20n+15}{40}\right) \frac{1}{5^n}$

$S_n = \frac{225}{32} - \frac{15}{8} \cdot \left(\frac{8n^2+20n+15}{8}\right) \frac{1}{5^n}$

$S_n = \frac{225}{32} - \left(\frac{120n^2 + 300n + 225}{320}\right)\frac{1}{5^n}$

$S_n = \frac{225}{32} - \left(\frac{24n^2 + 60n + 45}{64}\right)\frac{1}{5^n}$

3. 最終的な答え

$S_n = \frac{225}{32} - \left(\frac{24n^2 + 60n + 45}{64}\right) \frac{1}{5^n}$