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

The sum from n=0 to infinity of 4/(2^n) is 8

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

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

8

Solution steps

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

Apply the constant multiplication rule:

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

= 4 \cdot n = 0 \sum \infty \frac{1}{2 ^{n}}

Apply Series Geometric Test:

2

= 4 \cdot 2

Simplify

= 8

f (x) = cos (x)

(- 18 - 20) (- 18 - 15)

\int (\frac{x ^{6} - 5 x}{x ^{2}}) d x

x \to 0 lim (\frac{x}{sin ( \frac{x}{2} )})

\frac{d}{d x} (\frac{- 3 x ^{4} - 5 x ^{3} + 2}{- 5 x ^{4}})

What is the sum from n=0 to infinity of 4/(2^n) ?
The sum from n=0 to infinity of 4/(2^n) is 8