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人気のある微分積分 >

sum from k=1 to infinity of 2^{-k}

解

k = 1 \sum \infty 2^{- k}

解

1

解答ステップ

k = 1 \sum \infty 2^{- k}

簡素化

2^{- k} : \frac{1}{2 ^{k}}

= k = 1 \sum \infty \frac{1}{2 ^{k}}

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

= k = 0 \sum \infty \frac{1}{2 ^{k}} - \frac{1}{2 ^{0}}

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

= 2 - \frac{1}{2 ^{0}}

簡素化

= 1

人気の例

sum from k=5 to infinity of (e/pi)^{k-1}

k = 5 \sum \infty (\frac{e}{π})^{k - 1}

sum from k=4 to infinity of 1/((k-1)^4)

k = 4 \sum \infty \frac{1}{( k - 1 ) ^{4}}

sum from n=0 to infinity of 6(2/3)^n

n = 0 \sum \infty 6 (\frac{2}{3})^{n}

sum from n=0 to infinity of 2(1/2)^n

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

sum from n=1 to infinity of (3/4)^{n-1}

n = 1 \sum \infty (\frac{3}{4})^{n - 1}