sum from n=1 to infinity of (3^n)/(7^n)

解

n = 1 \sum \infty \frac{3 ^{n}}{7 ^{n}}

\frac{3}{4}

十進法表記

0.75

解答ステップ

n = 1 \sum \infty \frac{3 ^{n}}{7 ^{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{3}{7})^{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{3}{7})^{n} - (\frac{3}{7})^{0}

n = 0 \sum \infty (\frac{3}{7})^{n} = \frac{7}{4}

= \frac{7}{4} - (\frac{3}{7})^{0}

簡素化

\frac{7}{4} - (\frac{3}{7})^{0} : \frac{3}{4}

= \frac{3}{4}