sum from k=2 to infinity of 4/(7^k)

解

k = 2 \sum \infty \frac{4}{7 ^{k}}

\frac{2}{21}

十進法表記

0.09523 \dots

解答ステップ

k = 2 \sum \infty \frac{4}{7 ^{k}}

定数乗算の規則を適用する:

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

= 4 \cdot k = 2 \sum \infty \frac{1}{7 ^{k}}

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

= 4 (k = 0 \sum \infty \frac{1}{7 ^{k}} - \frac{1}{7 ^{0}} - \frac{1}{7 ^{1}})

k = 0 \sum \infty \frac{1}{7 ^{k}} = \frac{7}{6}

= 4 (\frac{7}{6} - \frac{1}{7 ^{0}} - \frac{1}{7 ^{1}})

簡素化

4 (\frac{7}{6} - \frac{1}{7 ^{0}} - \frac{1}{7 ^{1}}) : \frac{2}{21}

= \frac{2}{21}