sum from n=1 to infinity of (5)^{-n}

解答

n = 1 \sum \infty (5)^{- n}

\frac{1}{4}

十进制

0.25

求解步骤

n = 1 \sum \infty 5^{- n}

化简

5^{- n} : \frac{1}{5 ^{n}}

= n = 1 \sum \infty \frac{1}{5 ^{n}}

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

= n = 0 \sum \infty \frac{1}{5 ^{n}} - \frac{1}{5 ^{0}}

n = 0 \sum \infty \frac{1}{5 ^{n}} = \frac{5}{4}

= \frac{5}{4} - \frac{1}{5 ^{0}}

化简

\frac{5}{4} - \frac{1}{5 ^{0}} : \frac{1}{4}

= \frac{1}{4}

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