sum from n=1 to infinity of (2^n)/(3^n)

解

n = 1 \sum \infty \frac{2 ^{n}}{3 ^{n}}

2

解答ステップ

n = 1 \sum \infty \frac{2 ^{n}}{3 ^{n}}

指数の規則を適用する:

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

\frac{2 ^{n}}{3 ^{n}} = (\frac{2}{3})^{n}

= n = 1 \sum \infty (\frac{2}{3})^{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{2}{3})^{n} - (\frac{2}{3})^{0}

n = 0 \sum \infty (\frac{2}{3})^{n} = 3

= 3 - (\frac{2}{3})^{0}

簡素化

= 2