sum from n=1 to infinity of 2/25*(1/5)^n

解

n = 1 \sum \infty \frac{2}{25} \cdot (\frac{1}{5})^{n}

\frac{1}{50}

十進法表記

0.02

解答ステップ

n = 1 \sum \infty \frac{2}{25} (\frac{1}{5})^{n}

定数乗算の規則を適用する:

\sum c \cdot a_{n} = c \cdot \sum a_{n}

= \frac{2}{25} \cdot n = 1 \sum \infty (\frac{1}{5})^{n}

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

= \frac{2}{25} (n = 0 \sum \infty (\frac{1}{5})^{n} - (\frac{1}{5})^{0})

n = 0 \sum \infty (\frac{1}{5})^{n} = \frac{5}{4}

= \frac{2}{25} (\frac{5}{4} - (\frac{1}{5})^{0})

簡素化

\frac{2}{25} (\frac{5}{4} - (\frac{1}{5})^{0}) : \frac{1}{50}

= \frac{1}{50}