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人気のある微分積分 >

sum from n=2 to infinity of (0.2)^n

解

n = 2 \sum \infty (0.2)^{n}

解

0.05

解答ステップ

n = 2 \sum \infty 0. 2^{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. 2^{n} - 0. 2^{0} - 0. 2^{1}

n = 0 \sum \infty 0. 2^{n} = 1.25

= 1.25 - 0. 2^{0} - 0. 2^{1}

簡素化

= 0.05

人気の例

sum from n=0 to infinity of cos(4n)

n = 0 \sum \infty cos (4 n)

sum from n=0 to infinity of e^{-10n}

n = 0 \sum \infty e^{- 10 n}

sum from k=1 to infinity of (2^k)/((k)!)

k = 1 \sum \infty \frac{2 ^{k}}{( k ) !}

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

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

sum from n=1 to infinity of (6^n)/(n+1)

n = 1 \sum \infty \frac{6 ^{n}}{n + 1}