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

The sum from n=0 to infinity of 4(3/4)^n is 16

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

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

16

Solution steps

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

Apply the constant multiplication rule:

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

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

Apply Series Geometric Test:

4

= 4 \cdot 4

Simplify

= 16

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

\frac{2 s + 1}{s ^{3}}

h (x) = \frac{x ^{2} + 4 x + 1}{3 x + 5}

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

\int 2 x \cdot e^{2 x} d x

What is the sum from n=0 to infinity of 4(3/4)^n ?
The sum from n=0 to infinity of 4(3/4)^n is 16