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人気のある微分積分 >

sum from n=0 to infinity of 3^n*7^{-n}

解

n = 0 \sum \infty 3^{n} \cdot 7^{- n}

解

\frac{7}{4}

+1

十進法表記

1.75

解答ステップ

n = 0 \sum \infty 3^{n} \cdot 7^{- n}

簡素化

7^{- n} : \frac{1}{7 ^{n}}

= n = 0 \sum \infty 3^{n} \cdot \frac{1}{7 ^{n}}

簡素化

3^{n} \cdot \frac{1}{7 ^{n}} : (\frac{3}{7})^{n}

= n = 0 \sum \infty (\frac{3}{7})^{n}

級数幾何テストを適用する:

\frac{7}{4}

= \frac{7}{4}

人気の例

sum from n=1 to infinity of n/(nsqrt(n))

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

sum from n=1 to infinity of 2/(3n+7)

n = 1 \sum \infty \frac{2}{3 n + 7}

sum from n=0 to infinity of ((10pi)/e)^n

n = 0 \sum \infty (\frac{10 π}{e})^{n}

sum from k=1 to infinity of (k!)/(105^k)

k = 1 \sum \infty \frac{k !}{10 5 ^{k}}

sum from n=0 to infinity of \sqrt[n]{3}

n = 0 \sum \infty n 3