What is the sum from n=0 to infinity of 2*(4/5)^n ?

The sum from n=0 to infinity of 2*(4/5)^n is 10

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

sum from n=0 to infinity of 2*(4/5)^n

n = 0 \sum \infty 2 \cdot (\frac{4}{5})^{n}

10

Solution steps

n = 0 \sum \infty 2 (\frac{4}{5})^{n}

Apply the constant multiplication rule:

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

= 2 \cdot n = 0 \sum \infty (\frac{4}{5})^{n}

Apply Series Geometric Test:

5

= 2 \cdot 5

Simplify

= 10

y = 9 - x^{2}, y = - 8 x

\frac{2 x}{x ^{2} - 4}

\int \frac{5 x}{5 x ^{2} + 4 y ^{2}} d x

\int - 8 x^{- 9} d x

What is the sum from n=0 to infinity of 2*(4/5)^n ?
The sum from n=0 to infinity of 2*(4/5)^n is 10