sum from k=1 to infinity of 8(1/5)^k

解

k = 1 \sum \infty 8 (\frac{1}{5})^{k}

2

解答ステップ

k = 1 \sum \infty 8 (\frac{1}{5})^{k}

定数乗算の規則を適用する:

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

= 8 \cdot k = 1 \sum \infty (\frac{1}{5})^{k}

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

= 8 (k = 0 \sum \infty (\frac{1}{5})^{k} - (\frac{1}{5})^{0})

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

= 8 (\frac{5}{4} - (\frac{1}{5})^{0})

簡素化

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

= 2