sum from n=1 to infinity of (8^n)/(9^n)

解

n = 1 \sum \infty \frac{8 ^{n}}{9 ^{n}}

8

解答ステップ

n = 1 \sum \infty \frac{8 ^{n}}{9 ^{n}}

指数の規則を適用する:

\frac{a ^{c}}{b ^{c}} = (\frac{a}{b})^{c}

= n = 1 \sum \infty (\frac{8}{9})^{n}

n = k \sum \infty (a_{n}) = n = 0 \sum \infty (a_{n}) - a_{0} - a_{1} - \dots a_{k - 1}

= n = 0 \sum \infty (\frac{8}{9})^{n} - (\frac{8}{9})^{0}

n = 0 \sum \infty (\frac{8}{9})^{n} = 9

= 9 - (\frac{8}{9})^{0}

簡素化

= 8