ja

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

人気のある微分積分 >

sum from n=0 to infinity of 0.52(0.01)^n

解

n = 0 \sum \infty 0.52 (0.01)^{n}

解

0.52525 \dots

解答ステップ

n = 0 \sum \infty 0.52 \cdot 0.0 1^{n}

定数乗算の規則を適用する:

\sum c \cdot a_{n} = c \cdot \sum a_{n}

= 0.52 \cdot n = 0 \sum \infty 0.0 1^{n}

級数幾何テストを適用する:

1.01010 \dots

= 0.52 \cdot 1.01010 \dots

簡素化

= 0.52525 \dots

人気の例

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

n = 1 \sum \infty 2 + \frac{cos ( n )}{n}

sum from n=0 to infinity of (13/3)^n

n = 0 \sum \infty (\frac{13}{3})^{n}

sum from n=0 to infinity of (-1)^n+1

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

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

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

sum from n=0 to infinity of ln(3+n)

n = 0 \sum \infty ln (3 + n)