sum from n=1 to infinity of (4/10)^n*3

解

n = 1 \sum \infty (\frac{4}{10})^{n} \cdot 3

2

解答ステップ

n = 1 \sum \infty (\frac{4}{10})^{n} \cdot 3

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{4}{10})^{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{4}{10})^{n} - (\frac{4}{10})^{0})

n = 0 \sum \infty (\frac{4}{10})^{n} = \frac{5}{3}

= 3 (\frac{5}{3} - (\frac{4}{10})^{0})

簡素化

3 (\frac{5}{3} - (\frac{4}{10})^{0}) : 2

= 2