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人気のある微分積分 >

sum from n=1 to infinity of (0.1)^n

解

n = 1 \sum \infty (0.1)^{n}

解

0.11111 \dots

解答ステップ

n = 1 \sum \infty 0. 1^{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 0. 1^{n} - 0. 1^{0}

n = 0 \sum \infty 0. 1^{n} = 1.11111 \dots

= 1.11111 \dots - 0. 1^{0}

簡素化

= 0.11111 \dots

人気の例

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