sum from n=2 to infinity of 7/(2^n)

解

n = 2 \sum \infty \frac{7}{2 ^{n}}

\frac{7}{2}

十進法表記

3.5

解答ステップ

n = 2 \sum \infty \frac{7}{2 ^{n}}

定数乗算の規則を適用する:

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

= 7 \cdot n = 2 \sum \infty \frac{1}{2 ^{n}}

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

= 7 (n = 0 \sum \infty \frac{1}{2 ^{n}} - \frac{1}{2 ^{0}} - \frac{1}{2 ^{1}})

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

= 7 (2 - \frac{1}{2 ^{0}} - \frac{1}{2 ^{1}})

簡素化

7 (2 - \frac{1}{2 ^{0}} - \frac{1}{2 ^{1}}) : \frac{7}{2}

= \frac{7}{2}