sum from n=1 to infinity of-1/(6^n)

解

n = 1 \sum \infty - \frac{1}{6 ^{n}}

- \frac{1}{5}

十進法表記

- 0.2

解答ステップ

n = 1 \sum \infty - \frac{1}{6 ^{n}}

定数乗算の規則を適用する:

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

= - n = 1 \sum \infty \frac{1}{6 ^{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{1}{6 ^{n}} - \frac{1}{6 ^{0}})

n = 0 \sum \infty \frac{1}{6 ^{n}} = \frac{6}{5}

= - (\frac{6}{5} - \frac{1}{6 ^{0}})

簡素化

- (\frac{6}{5} - \frac{1}{6 ^{0}}) : - \frac{1}{5}

= - \frac{1}{5}