What is the sum from n=0 to infinity of 9/(2^n) ?

The sum from n=0 to infinity of 9/(2^n) is 18

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sum from n=0 to infinity of 9/(2^n)

n = 0 \sum \infty \frac{9}{2 ^{n}}

18

Solution steps

n = 0 \sum \infty \frac{9}{2 ^{n}}

Apply the constant multiplication rule:

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

= 9 \cdot n = 0 \sum \infty \frac{1}{2 ^{n}}

Apply Series Geometric Test:

2

= 9 \cdot 2

Simplify

= 18

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

n = 1 \sum \infty \frac{n}{n ^{2} + 16}

k = 1 \sum \infty (\frac{π}{e})^{- k}

n = 1 \sum \infty \frac{n}{cos ( n )}

n = 1 \sum \infty \frac{1}{n ^{2} - 4 n + 5}

What is the sum from n=0 to infinity of 9/(2^n) ?
The sum from n=0 to infinity of 9/(2^n) is 18