What is the sum from n=0 to infinity of (3+2n)/(2^n) ?

The sum from n=0 to infinity of (3+2n)/(2^n) is converges

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sum from n=0 to infinity of (3+2n)/(2^n)

n = 0 \sum \infty \frac{3 + 2 n}{2 ^{n}}

converges

Solution steps

n = 0 \sum \infty \frac{3 + 2 n}{2 ^{n}}

Expand

\frac{3 + 2 n}{2 ^{n}} : \frac{3}{2 ^{n}} + 2^{1 - n} n

= n = 0 \sum \infty \frac{3}{2 ^{n}} + 2^{1 - n} n

Apply the Sum Rule:

\sum a_{n} + b_{n} = \sum a_{n} + \sum b_{n}

= n = 0 \sum \infty \frac{3}{2 ^{n}} + n = 0 \sum \infty 2^{1 - n} n

n = 0 \sum \infty \frac{3}{2 ^{n}} = 6

n = 0 \sum \infty 2^{1 - n} n =

converges

= 6 + converges

= converges

n = 0 \sum \infty \frac{1}{n + 13}

\int \frac{y}{( y ^{2} - 4 ) ^{\frac{2}{5}}} d y

f (x) = 12 x - x^{3}

\int - cos (n x) d x

\frac{d y}{d x} = \frac{x ^{3} + y ^{3}}{x y ^{2}}

What is the sum from n=0 to infinity of (3+2n)/(2^n) ?
The sum from n=0 to infinity of (3+2n)/(2^n) is converges