Sigma from k=1 to infinity of (-1)^k+1(4^k)(9^1-k)
How do I deal with that (-1)^k+1?
If the series is $\displaystyle \sum^\infty_{k=1}(-1)^{k+1}4^k\cdot9^{1-k}=-9\sum^\infty_{k=1}\left(-\frac{4}{9}\right)^k$ , then all you have to do is remember that $\displaystyle |q|<1\Longrightarrow \sum^\infty_{k=0}aq^k=\frac{a}{1-q}\,,\,\,\forall\,a\in\mathbb{R}$ (sum of an
infinite convergent geometric series)
Tonio