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Math Help - Help with series needed please

  1. #1
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    Help with series needed please

    Hello, can somebody please explain the intermediate steps I take to solve the following?

    Thanks!

    \displaystyle \frac{2}{M} \sum_{n=1}^{\frac{M}{2}} (2n-1)^2 A^2 =  \frac{M^2-1}{3} A^2
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by jayd1988 View Post
    Hello, can somebody please explain the intermediate steps I take to solve the following?

    Thanks!

    \displaystyle \frac{2}{M} \sum_{n=1}^{\frac{M}{2}} (2n-1)^2 A^2 =  \frac{M^2-1}{3} A^2
    First off, the A^2 can be pulled out. So we have \displaystyle\frac{2A^2}{M}\sum\limits_{n=1}^{\fra  c{M}{2}}(2n-1)^2. Take k=2n-1. Then we can rewrite the sum as \displaystyle\frac{2A^2}{M}\sum\limits_{k=1}^{M-1}k^2. Now apply the formula for \displaystyle\sum_{k=1}^n k^2 and simplify. From there, you should get the desired result.

    Does this make sense?
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  3. #3
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    Quote Originally Posted by Chris L T521 View Post
    First off, the A^2 can be pulled out. So we have \displaystyle\frac{2A^2}{M}\sum\limits_{n=1}^{\fra  c{M}{2}}(2n-1)^2. Take k=2n-1. Then we can rewrite the sum as \displaystyle\frac{2A^2}{M}\sum\limits_{k=1}^{M-1}k^2. Now apply the formula for \displaystyle\sum_{k=1}^n k^2 and simplify. From there, you should get the desired result.

    Does this make sense?
    Many thanks for your help. Sorry, which formula do I use for \displaystyle\sum_{k=1}^n k^2 almost there
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  4. #4
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by jayd1988 View Post
    Many thanks for your help. Sorry, which formula do I use for \displaystyle\sum_{k=1}^n k^2 almost there
    \displaystyle\sum\limits_{k=1}^n k^2=\frac{n(n+1)(2n+1)}{6}
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  5. #5
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    \displaystyle\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}
    Last edited by pickslides; November 9th 2010 at 03:04 PM.
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  6. #6
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    Thanks guys! Got it...
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