(1) a1=5a2=35a3=3⋅3⋅514=332⋅514a4=3⋅334⋅518=374⋅518a_1 = 5 \qquad a_2 = 3\sqrt{5} \qquad a_3 = 3\cdot\sqrt{3}\cdot 5^{\frac{1}{4}} = 3^{\frac{3}{2}}\cdot 5^{\frac{1}{4}} \qquad a_4 = 3\cdot 3^{\frac{3}{4}}\cdot 5^{\frac{1}{8}} = 3^{\frac{7}{4}}\cdot 5^{\frac{1}{8}}a1=5a2=35a3=3⋅3⋅541=323⋅541a4=3⋅343⋅581=347⋅581 a5=3⋅378⋅5116=3158⋅5116a_5 = 3\cdot 3^{\frac{7}{8}}\cdot 5^{\frac{1}{16}} = 3^{\frac{15}{8}}\cdot 5^{\frac{1}{16}}a5=3⋅387⋅5161=3815⋅5161 an=3α⋅5βα=2−(12)n−2β=(12)n−1a_n = 3^{\alpha}\cdot 5^{\beta} \qquad \alpha = 2 - \left(\frac{1}{2}\right)^{n-2} \qquad \beta = \left(\frac{1}{2}\right)^{n-1}an=3α⋅5βα=2−(21)n−2β=(21)n−1 (2) limn→∞3α⋅5β\displaystyle\lim_{n\to\infty} 3^{\alpha}\cdot 5^{\beta}n→∞lim3α⋅5β において limn→∞α=limn→∞{2−(12)n−2}=2−0=2limn→∞β=limn→∞(12)n−1=0これから\lim_{n\to\infty}\alpha = \lim_{n\to\infty}\left\{2 - \left(\frac{1}{2}\right)^{n-2}\right\} = 2 - 0 = 2 \qquad \lim_{n\to\infty}\beta = \lim_{n\to\infty}\left(\frac{1}{2}\right)^{n-1} = 0 \quad \text{これから}limn→∞α=limn→∞{2−(21)n−2}=2−0=2limn→∞β=limn→∞(21)n−1=0これから limn→∞3α⋅5β=32⋅50=9\lim_{n\to\infty} 3^{\alpha}\cdot 5^{\beta} = 3^2 \cdot 5^0 = 9limn→∞3α⋅5β=32⋅50=9