What is the sum from n=0 to infinity of 3/(4^n) ?

The sum from n=0 to infinity of 3/(4^n) is 4

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 3/(4^n)

n = 0 \sum \infty \frac{3}{4 ^{n}}

4

Solution steps

n = 0 \sum \infty \frac{3}{4 ^{n}}

Apply the constant multiplication rule:

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

= 3 \cdot n = 0 \sum \infty \frac{1}{4 ^{n}}

Apply Series Geometric Test:

\frac{4}{3}

= 3 \cdot \frac{4}{3}

Simplify

3 \cdot \frac{4}{3} : 4

= 4

\frac{d}{d x} (x^{2} π)

\int \frac{( 1 + x ^{2} )}{x} d x

\frac{3 x}{8 - cot ( x )}

f (x) = 5^{- \frac{x}{2}} sin (2 x)

x \to 0 lim (x^{4} + 2 e^{x} - 2)

What is the sum from n=0 to infinity of 3/(4^n) ?
The sum from n=0 to infinity of 3/(4^n) is 4