# Math Help - Series Help

1. ## Series Help

Sigma from k=1 to infinity of (-1)^k+1(4^k)(9^1-k)

How do I deal with that (-1)^k+1?

2. Originally Posted by Simon777
Sigma from k=1 to infinity of (-1)^k+1(4^k)(9^1-k)

How do I deal with that (-1)^k+1?
$(-1)^{k+1} \frac{4^k}{9^{k-1}} =$

$-9 \cdot (-1)^k \frac{4^k}{9^{k}} =$

$-9 \left(-\frac{4}{9}\right)^k$

3. Originally Posted by Simon777
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 $\sum^\infty_{k=1}(-1)^{k+1}4^k\cdot9^{1-k}=-9\sum^\infty_{k=1}\left(-\frac{4}{9}\right)^k$ , then all you have to do is remember that $|q|<1\Longrightarrow \sum^\infty_{k=0}aq^k=\frac{a}{1-q}\,,\,\,\forall\,a\in\mathbb{R}$ (sum of an

infinite convergent geometric series)

Tonio