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}

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}