What is the sum from n=1 to infinity of 4/(2^n) ?

The sum from n=1 to infinity of 4/(2^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=1 to infinity of 4/(2^n)

n = 1 \sum \infty \frac{4}{2 ^{n}}

4

Solution steps

n = 1 \sum \infty \frac{4}{2 ^{n}}

Apply the constant multiplication rule:

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

= 4 \cdot n = 1 \sum \infty \frac{1}{2 ^{n}}

n = k \sum \infty (a_{n}) = n = 0 \sum \infty (a_{n}) - a_{0} - a_{1} - \dots a_{k - 1}

= 4 (n = 0 \sum \infty \frac{1}{2 ^{n}} - \frac{1}{2 ^{0}})

n = 0 \sum \infty \frac{1}{2 ^{n}} = 2

= 4 (2 - \frac{1}{2 ^{0}})

Simplify

= 4

\int_{2}^{- 1} x^{2} + 1 d x

\int \frac{x}{1 + x ^{4}} d x

\frac{d}{d x} (5 x y^{2})

\frac{( x - 13 )}{( ( x ^{5} ) - ( 135 ) )}

\frac{d x}{d t} = \frac{( e ^{t} - e ^{- t} )}{x + 3}

What is the sum from n=1 to infinity of 4/(2^n) ?
The sum from n=1 to infinity of 4/(2^n) is 4