What is the sum from k=1 to infinity of (5+(-2)^k)/(4^{k+1)} ?

The sum from k=1 to infinity of (5+(-2)^k)/(4^{k+1)} is converges

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sum from k=1 to infinity of (5+(-2)^k)/(4^{k+1)}

k = 1 \sum \infty \frac{5 + ( - 2 ) ^{k}}{4 ^{k + 1}}

converges

Solution steps

k = 1 \sum \infty \frac{5 + ( - 2 ) ^{k}}{4 ^{k + 1}}

Expand

\frac{5 + ( - 2 ) ^{k}}{4 ^{k + 1}} : \frac{5}{4 ^{k + 1}} + \frac{( - 2 ) ^{k}}{4 ^{k + 1}}

= k = 1 \sum \infty \frac{5}{4 ^{k + 1}} + \frac{( - 2 ) ^{k}}{4 ^{k + 1}}

Apply the Sum Rule:

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

= k = 1 \sum \infty \frac{5}{4 ^{k + 1}} + k = 1 \sum \infty \frac{( - 2 ) ^{k}}{4 ^{k + 1}}

k = 1 \sum \infty \frac{5}{4 ^{k + 1}} =

converges

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

converges

= converges + converges

= converges

\int \frac{sin ^{7} ( 12 x )}{cos ^{4} ( 12 x )} d x

f (x) = ln (\frac{7 ( x + 3 )}{x})

\int x^{2} e^{- \frac{x}{3}} d x

\int_{1}^{5} \frac{1}{4 x + 5} d x

y^{^{''}} - 2 y = 0

What is the sum from k=1 to infinity of (5+(-2)^k)/(4^{k+1)} ?
The sum from k=1 to infinity of (5+(-2)^k)/(4^{k+1)} is converges