What is the sum from n=0 to infinity of 6/(4^n) ?

The sum from n=0 to infinity of 6/(4^n) is 8

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 6/(4^n)

n = 0 \sum \infty \frac{6}{4 ^{n}}

8

Solution steps

n = 0 \sum \infty \frac{6}{4 ^{n}}

Apply the constant multiplication rule:

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

= 6 \cdot n = 0 \sum \infty \frac{1}{4 ^{n}}

Apply Series Geometric Test:

\frac{4}{3}

= 6 \cdot \frac{4}{3}

Simplify

6 \cdot \frac{4}{3} : 8

= 8

\int x e^{x + y} d x

\frac{d}{d x} (x^{3} + x - 2)

x \to 2 lim (\frac{5}{x} - 8)

\int - x^{2} + 2 x + 3 d x

\frac{d y}{d x} = 4 x^{3} y - 6 x^{2} y

What is the sum from n=0 to infinity of 6/(4^n) ?
The sum from n=0 to infinity of 6/(4^n) is 8