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、$S_n$ を求めます。

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、$S_n$ を求めます。

2. 解き方の手順

まず、与えられた式を整理します。

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

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

$\frac{4}{75}S_n = \frac{5}{8} - \frac{8n+10}{16\cdot5^n} - \frac{1}{4} + \frac{1}{4\cdot5^n} - \frac{n^2}{5^{n+1}}$

$\frac{4}{75}S_n = \frac{5}{8} - \frac{2}{8} - \frac{8n+10}{16\cdot5^n} + \frac{4}{16\cdot5^n} - \frac{n^2}{5\cdot5^n}$

$\frac{4}{75}S_n = \frac{3}{8} - \frac{8n+6}{16\cdot5^n} - \frac{n^2}{5^{n+1}}$

$\frac{4}{75}S_n = \frac{3}{8} - \frac{4n+3}{8\cdot5^n} - \frac{n^2}{5^{n+1}}$

$\frac{4}{75}S_n = \frac{3}{8} - \frac{5(4n+3)}{8\cdot5^{n+1}} - \frac{8n^2}{8\cdot5^{n+1}}$

$\frac{4}{75}S_n = \frac{3}{8} - \frac{20n+15+8n^2}{8\cdot5^{n+1}}$

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

$S_n = \frac{75}{4} \cdot \frac{3}{8} - \frac{75(8n^2+20n+15)}{32 \cdot 5^{n+1}}$

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

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

$S_n = \frac{225}{32} - \frac{120n^2+300n+225}{32 \cdot 5^{n}}$

$S_n = \frac{225\cdot 5^n - (120n^2+300n+225)}{32 \cdot 5^n}$

3. 最終的な答え

$S_n = \frac{225 \cdot 5^n - (120n^2 + 300n + 225)}{32 \cdot 5^n}$

または

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