Question

Calculate the serie

\[\sum_{n=0}^{\infty}(2^{-n}-(\frac{-3}{4})^{n+2})\]

Solution

\[\sum_{n=0}^{\infty}(2^{-n})-\sum_{n=0}^{\infty}(\frac{-3}{4})^{n+2} = \]

\[\sum_{n=0}^{\infty}((\frac{1}{2})^{n})-\sum_{n=0}^{\infty}(\frac{-3}{4})^{n}.(\frac{-3}{4}^{2}) = \]
\[\sum_{n=0}^{\infty} \frac{1}{2}^{n} -\sum_{n=0}^{\infty} \frac{-3}{4}^{n}.\sum_{n=0}^{\infty}\frac{9}{16} = \frac{9}{16}
\]
can you check it out if the solution correct is?