sum from n=1 to infinity of 8*(1/3)^n

解

n = 1 \sum \infty 8 \cdot (\frac{1}{3})^{n}

4

解答ステップ

n = 1 \sum \infty 8 (\frac{1}{3})^{n}

定数乗算の規則を適用する:

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

= 8 \cdot n = 1 \sum \infty (\frac{1}{3})^{n}

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

= 8 (n = 0 \sum \infty (\frac{1}{3})^{n} - (\frac{1}{3})^{0})

n = 0 \sum \infty (\frac{1}{3})^{n} = \frac{3}{2}

= 8 (\frac{3}{2} - (\frac{1}{3})^{0})

簡素化

8 (\frac{3}{2} - (\frac{1}{3})^{0}) : 4

= 4