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

The sum from n=0 to infinity of 8/(4^n) is 32/3

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

n = 0 \sum \infty \frac{8}{4 ^{n}}

\frac{32}{3}

Solution steps

n = 0 \sum \infty \frac{8}{4 ^{n}}

Apply the constant multiplication rule:

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

= 8 \cdot n = 0 \sum \infty \frac{1}{4 ^{n}}

Apply Series Geometric Test:

\frac{4}{3}

= 8 \cdot \frac{4}{3}

Simplify

8 \cdot \frac{4}{3} : \frac{32}{3}

= \frac{32}{3}

\frac{\partial}{\partial x} (- sin (x^{2} y) x^{2})

\int t \cdot \frac{1}{e ^{t}} d t

\int (- 2 sec (x) tan (x) - 3 sec^{2} (x)) d x

5 t \cdot \frac{d y}{d t} + 3 y = t

\frac{d x}{d t} = \frac{( 2 t ^{2} + x ^{2} )}{t x}

What is the sum from n=0 to infinity of 8/(4^n) ?
The sum from n=0 to infinity of 8/(4^n) is 32/3