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Math Help - Help with evaluating a sum

  1. #1
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    Help with evaluating a sum

    Hello everyone,

    I'm not sure if this is the best category for my post, but i will try asking anyways. I've spent quite a bit of time trying to figure it out.

    Basically, the problem involves calculating the output of a system to a given input and impulse response using convolution. That's probably the hardest part of the question, which I am able to do.

    I get the output y[n] = summation(from k=0 to k=n) ((1/2)^k)*(1/3)^(n-k)
    So i know that (1/3)^n can be brought outside the summation, and that
    ((1/2)^k)*(1/3)^(-k) = (3/2)^k, so the question can be simplified to

    y[n]=((1/3)^n)*summation(k=0 to k=n) (3/2)^k

    I know the formula for evaluating a sum from 0 to n, but for some reason i can't simplify it to the answer in the textbook. The answer i need is the following:

    y[n]=2[((3/2)^(n+1))-1]

    Any help is greatly appreciated.

    The question can be seen here: http://www.site.uottawa.ca/~jpyao/co..._Fall_2004.pdf
    Question #4
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  2. #2
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    Re: Help with evaluating a sum

    \left(\frac{1}{3}\right)^n \sum_{k=0}^n \left(\frac{3}{2}\right)^k

    sum of the finite geometric series ...

    S_n = \frac{a_0 - r^{n+1}}{1 - r}

    S_n = \frac{1 - \left(\frac{3}{2}\right)^{n+1}}{1 - \left(\frac{3}{2}\right)}

    S_n = \frac{1 - \left(\frac{3}{2}\right)^{n+1}}{-\frac{1}{2}}

    S_n = 2\left[\left(\frac{3}{2}\right)^{n+1} - 1\right]

    so ...

    \left(\frac{1}{3}\right)^n \sum_{k=0}^n \left(\frac{3}{2}\right)^k = \left(\frac{1}{3}\right)^n \cdot 2\left[\left(\frac{3}{2}\right)^{n+1} - 1\right]


    did the book leave off the \left(\frac{1}{3}\right)^n or did you?
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  3. #3
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    Re: Help with evaluating a sum

    Wow, amazing how simple that was. I was way overcomplicating it.

    The book left off the (1/3)^n apparently. Thanks for clarifying it for me. Still can't believe i spent over an hour trying to figure it out!
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