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Math Help - Summation proving 2

  1. #1
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    Summation proving 2

    Making use of the method of differences, prove that \sum_{n = 1}^{k}\frac{n^2+6n+4}{(n+2)!}=3-\frac{1}{(k+1)!}-\frac{4}{(k+2)!}

    \sum_{n = 1}^{k}\frac{n^2+6n+4}{(n+2)!}=\sum_{n=1}^{k}\frac{  (n+2)(n+1)+3(n+2)-4}{(n+2)!}

    =\sum_{n = 1}^{k}\frac{(n+2)(n+1)}{(n+2)!}+\frac{3(n+2)}{(n+2  )!}-\frac{4}{(n+2)!}

    =\sum_{n = 1}^{k}\frac{(n+2)(n+1)}{(n+2)(n+1)n!}+\frac{3(n+2)  }{(n+2)(n+1)!}-\frac{4}{(n+2)!}

    =\sum_{n = 1}^{k}\frac{1}{n!}+\frac{3}{(n+1)!}-\frac{4}{n+2)!}
    Last edited by Punch; May 21st 2011 at 06:38 AM.
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  2. #2
    Super Member TheChaz's Avatar
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    Your first equality is false.
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  3. #3
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    I missed out something, the first part of the question asked to verify that n^2+6n+4=(n+2)(n+1)+3(n+2)-4, so it should be true
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  4. #4
    Super Member TheChaz's Avatar
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    Quote Originally Posted by Punch View Post
    I missed out something, the first part of the question asked to verify that n^2+6n+4=(n+2)(n+1)+3(n+2)-4, so it should be true
    I was referring to the denominator.
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  5. #5
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    Quote Originally Posted by TheChaz View Post
    I was referring to the denominator.
    sorry it was a typo, i corrected it
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