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Math Help - Series Help

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    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?
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    Quote Originally Posted by Simon777 View Post
    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
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  3. #3
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    Quote Originally Posted by Simon777 View Post
    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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