sum from n=1 to infinity of 15(1/100)^n

解

n = 1 \sum \infty 15 (\frac{1}{100})^{n}

\frac{5}{33}

十進法表記

0.15151 \dots

解答ステップ

n = 1 \sum \infty 15 (\frac{1}{100})^{n}

定数乗算の規則を適用する:

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

= 15 \cdot n = 1 \sum \infty (\frac{1}{100})^{n}

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

= 15 (n = 0 \sum \infty (\frac{1}{100})^{n} - (\frac{1}{100})^{0})

n = 0 \sum \infty (\frac{1}{100})^{n} = \frac{100}{99}

= 15 (\frac{100}{99} - (\frac{1}{100})^{0})

簡素化

15 (\frac{100}{99} - (\frac{1}{100})^{0}) : \frac{5}{33}

= \frac{5}{33}