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人気のある微分積分 >

sum from n=1 to infinity of 1/(4^n)*2^n

解

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

解

1

解答ステップ

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

簡素化

\frac{1}{4 ^{n}} \cdot 2^{n} : (\frac{1}{2})^{n}

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

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

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

簡素化

= 1

人気の例

sum from n=1 to infinity of 4/(2^{n+2)}

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

sum from n=1 to infinity of (n!)/(112^n)

n = 1 \sum \infty \frac{n !}{11 2 ^{n}}

sum from n=1 to infinity of 3/(n^{5/2)}

n = 1 \sum \infty \frac{3}{n ^{\frac{5}{2}}}

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

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

sum from n=1 to infinity of n!(-e)^{-4n}

n = 1 \sum \infty n! (- e)^{- 4 n}