sum from n=2 to infinity of 8/(4^{2n)}

解

n = 2 \sum \infty \frac{8}{4 ^{2 n}}

\frac{1}{30}

十進法表記

0.03333 \dots

解答ステップ

n = 2 \sum \infty \frac{8}{4 ^{2 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 = 2 \sum \infty \frac{1}{4 ^{2 n}}

4^{2 n} = 1 6^{n}

= 8 \cdot n = 2 \sum \infty \frac{1}{1 6 ^{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}{1 6 ^{n}} - \frac{1}{1 6 ^{0}} - \frac{1}{1 6 ^{1}})

n = 0 \sum \infty \frac{1}{1 6 ^{n}} = \frac{16}{15}

= 8 (\frac{16}{15} - \frac{1}{1 6 ^{0}} - \frac{1}{1 6 ^{1}})

簡素化

8 (\frac{16}{15} - \frac{1}{1 6 ^{0}} - \frac{1}{1 6 ^{1}}) : \frac{1}{30}

= \frac{1}{30}