Series Help

**Simon777** · April 18th 2010, 05:58 PM

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

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

**skeeter** · April 18th 2010, 06:16 PM

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?

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

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

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

**tonio** · April 18th 2010, 06:20 PM

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

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