What is the sum from n=1 to infinity of (-2+3^n)/(2^{2n)} ?

The sum from n=1 to infinity of (-2+3^n)/(2^{2n)} is 7/3

Popular >

sum from n=1 to infinity of (-2+3^n)/(2^{2n)}

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

\frac{7}{3}

Solution steps

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

Expand

\frac{- 2 + 3 ^{n}}{2 ^{2 n}} : - \frac{2}{4 ^{n}} + (\frac{3}{4})^{n}

= n = 1 \sum \infty - \frac{2}{4 ^{n}} + (\frac{3}{4})^{n}

Apply the Sum Rule:

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

= - n = 1 \sum \infty \frac{2}{4 ^{n}} + n = 1 \sum \infty (\frac{3}{4})^{n}

n = 1 \sum \infty \frac{2}{4 ^{n}} = \frac{2}{3}

n = 1 \sum \infty (\frac{3}{4})^{n} = 3

= - \frac{2}{3} + 3

Simplify

- \frac{2}{3} + 3 : \frac{7}{3}

= \frac{7}{3}

n = 1 \sum \infty n 19

\frac{d}{d θ} (\frac{2}{6 cos ( θ ) + 7 sin ( θ )})

\int_{0}^{3} (\frac{1}{x ^{2}}) d x

x \to 2 lim (x^{2} - 8)

\frac{d}{d x} (- \frac{1}{4} x^{- \frac{3}{2}})

What is the sum from n=1 to infinity of (-2+3^n)/(2^{2n)} ?
The sum from n=1 to infinity of (-2+3^n)/(2^{2n)} is 7/3