What is the sum from n=0 to infinity of 20(1/4)^n ?

The sum from n=0 to infinity of 20(1/4)^n is 80/3

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sum from n=0 to infinity of 20(1/4)^n

n = 0 \sum \infty 20 (\frac{1}{4})^{n}

\frac{80}{3}

Solution steps

n = 0 \sum \infty 20 (\frac{1}{4})^{n}

Apply the constant multiplication rule:

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

= 20 \cdot n = 0 \sum \infty (\frac{1}{4})^{n}

Apply Series Geometric Test:

\frac{4}{3}

= 20 \cdot \frac{4}{3}

Simplify

20 \cdot \frac{4}{3} : \frac{80}{3}

= \frac{80}{3}

k = 2 \sum \infty (\frac{1}{3})^{2 k}

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

n = 1 \sum \infty \frac{5}{n ^{2} + 1}

k = 2 \sum \infty 9 e^{- 2 k}

k = 2 \sum \infty (\frac{1}{3})^{3 k}

What is the sum from n=0 to infinity of 20(1/4)^n ?
The sum from n=0 to infinity of 20(1/4)^n is 80/3