sum from n=3 to infinity of 2/(5^n)

解

n = 3 \sum \infty \frac{2}{5 ^{n}}

\frac{1}{50}

十進法表記

0.02

解答ステップ

n = 3 \sum \infty \frac{2}{5 ^{n}}

定数乗算の規則を適用する:

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

= 2 \cdot n = 3 \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}

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

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

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

簡素化

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

= \frac{1}{50}