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Math Help - Summation Induction help

  1. #1
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    Question Summation Induction help

    Hello, I've read everything I can on induction for the most part it makes sense, but I'm having trouble with this summation problem:

    SUM from i = 1 to n: (-1)^(i+1)(i^2) = [ (-1)^(n+1) n(n+1) ] / 2

    (I dont know latex...yet)

    Basis step:
    1 = 1 check

    Inductive step:
    plugging in n+1 and trying to solve I can't seem to get anywhere...
    Please point me in the right direction


    Thank-you for any help!!
    jeremywho
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  2. #2
    Junior Member
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    Well, I'll assume you know how induction is done and how it works. It's just a matter (in this case) of calculation. In fact, suppose the equality holds for some n \in \mathbb N, that is:

    \sum_{i=1}^n (-1)^{i+1}i^2=(-1)^{n+1}\frac{n(n+1)}{2}

    You wish to prove it also holds for n+1, hence you want to prove that:

    \sum_{i=1}^{n+1} (-1)^{i+1}i^2=(-1)^{n+2}\frac{(n+1)(n+2)}{2}

    Simply observe that:

    \sum_{i=1}^{n+1} (-1)^{i+1}i^2=(-1)^{n+2}(n+1)^2+\sum_{i=1}^{n} (-1)^{i+1}i^2

    and recall that you are assuming the equality is true for n (which means you know the corresponding expression for that sum on the right side).

    Hope it helped
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  3. #3
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    Thank-you so much for the reply!!

    I understand we want to prove this:
    <br />
\sum_{i=1}^{n+1} (-1)^{i+1}i^2=(-1)^{n+2}\frac{(n+1)(n+2)}{2}<br />

    I don't understand how we get:

    <br />
(-1)^{n+2}(n+1)^2<br />

    in

    <br />
\sum_{i=1}^{n+1} (-1)^{i+1}i^2=(-1)^{n+2}(n+1)^2+\sum_{i=1}^{n} (-1)^{i+1}i^2<br />

    Maybe its just simple algebra I'm missing?

    Thank-you Again!!
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  4. #4
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    Lisbon
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    You just pick the last term of

    \sum_{i=1}^{n+1} (-1)^{i+1}i^2

    which is (-1)^{n+2}(n+1)^2 and write it independently. A straightforward, numerical example:

    \sum_{i=1}^{5}i = 1+2+3+4+5 = 5+(4+3+2+1) = 5+ (\sum_{i=1}^4i)

    The sums don't have the same upper bound! One ends in n+1 and the other in n.
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  5. #5
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    Got it!

    I finally got it! I just wanted to thank-you again for your help!
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