無限級数 $\sum_{n=1}^{\infty} \frac{2^n + 3^n}{5^n}$ の和を求める問題です。解析学無限級数等比級数級数の和2025/7/81. 問題の内容無限級数 ∑n=1∞2n+3n5n\sum_{n=1}^{\infty} \frac{2^n + 3^n}{5^n}∑n=1∞5n2n+3n の和を求める問題です。2. 解き方の手順与えられた無限級数を2つの無限等比級数に分解します。∑n=1∞2n+3n5n=∑n=1∞(2n5n+3n5n)=∑n=1∞(25)n+∑n=1∞(35)n\sum_{n=1}^{\infty} \frac{2^n + 3^n}{5^n} = \sum_{n=1}^{\infty} \left( \frac{2^n}{5^n} + \frac{3^n}{5^n} \right) = \sum_{n=1}^{\infty} \left(\frac{2}{5}\right)^n + \sum_{n=1}^{\infty} \left(\frac{3}{5}\right)^n∑n=1∞5n2n+3n=∑n=1∞(5n2n+5n3n)=∑n=1∞(52)n+∑n=1∞(53)n無限等比級数の公式を利用します。∑n=1∞arn−1=a1−r\sum_{n=1}^{\infty} ar^{n-1} = \frac{a}{1-r}∑n=1∞arn−1=1−ra (∣r∣<1|r| < 1∣r∣<1)または ∑n=1∞rn=r1−r\sum_{n=1}^{\infty} r^n = \frac{r}{1-r}∑n=1∞rn=1−rr (∣r∣<1|r| < 1∣r∣<1)∑n=1∞(25)n=251−25=2535=23\sum_{n=1}^{\infty} \left(\frac{2}{5}\right)^n = \frac{\frac{2}{5}}{1 - \frac{2}{5}} = \frac{\frac{2}{5}}{\frac{3}{5}} = \frac{2}{3}∑n=1∞(52)n=1−5252=5352=32∑n=1∞(35)n=351−35=3525=32\sum_{n=1}^{\infty} \left(\frac{3}{5}\right)^n = \frac{\frac{3}{5}}{1 - \frac{3}{5}} = \frac{\frac{3}{5}}{\frac{2}{5}} = \frac{3}{2}∑n=1∞(53)n=1−5353=5253=23したがって、∑n=1∞2n+3n5n=23+32=46+96=136\sum_{n=1}^{\infty} \frac{2^n + 3^n}{5^n} = \frac{2}{3} + \frac{3}{2} = \frac{4}{6} + \frac{9}{6} = \frac{13}{6}∑n=1∞5n2n+3n=32+23=64+69=6133. 最終的な答え136\frac{13}{6}613