sum from n=1 to infinity of 3*(1/5)^n

解

n = 1 \sum \infty 3 \cdot (\frac{1}{5})^{n}

\frac{3}{4}

十進法表記

0.75

解答ステップ

n = 1 \sum \infty 3 (\frac{1}{5})^{n}

定数乗算の規則を適用する:

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

= 3 \cdot n = 1 \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}

= 3 (n = 0 \sum \infty (\frac{1}{5})^{n} - (\frac{1}{5})^{0})

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

= 3 (\frac{5}{4} - (\frac{1}{5})^{0})

簡素化

3 (\frac{5}{4} - (\frac{1}{5})^{0}) : \frac{3}{4}

= \frac{3}{4}