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Math Help - proving this summation

  1. #1
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    proving this summation

    Prove that \sum^{n}_{r=1}[(r+2)(r+1)-3(r+1)+2]r!=n(n+1)!

    Thanks a lot .
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  2. #2
    MHF Contributor Amer's Avatar
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    Quote Originally Posted by thereddevils View Post
    Prove that \sum^{n}_{r=1}[(r+2)(r+1)-3(r+1)+2]r!=n(n+1)!

    Thanks a lot .
    let prove it by induction okay

    let P_n : \sum_{r=1}^{n} [(r+2)(r+1) - 3(r+1) +2]r! = n(n+1)!

    first see if n=1 and 2 true

    (1+2)(1+1)-3(2)+2 = 2 and

    (1)(2)! = 2 at 1 the statement is true try when n=2

    you will find it is true now suppose that n is true try n+1

    P_{n+1}:<br />
\sum_{r=1}^{n+1} [(r+2)(r+1)-3(r+1)+2]r! = (n+1)(n+2)! ......(*)

    \sum_{r=1}^{n+1}[(r+2)(r+1)-3(r+1)+2]r! = \underbrace{\sum_{r=1}^{n} [(r+2)(r+1)-3(r+1)+2]r! }_{n(n+1)!}+ [(n+3)(n+2)-3(n+2)+2](n+1)!

    \sum_{r=1}^{n+1}[(r+2)(r+1)-3(r+1)+2]r! = n(n+1)!+[(n+3)(n+2)-3(n+2)+2](n+1)!

    =(n+1)![n+(n+3)(n+2)-3(n+2)+2]

    =(n+1)![n+n^2+5n+6-3n-6+2]

    =(n+1)!(n^2+3n+2)

    =(n+1)!(n+1)(n+2)=(n+1)(n+2)! you begin with the RHS in (*) and finished with LHS in the (*) formula

    so P_{n+1} is true the proof end
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  3. #3
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    Hello thereddevils
    Quote Originally Posted by thereddevils View Post
    Prove that \sum^{n}_{r=1}[(r+2)(r+1)-3(r+1)+2]r!=n(n+1)!

    Thanks a lot .
    It's much easier if you simplify the expression first!

    (r+2)(r+1)-3(r+1)+2 = r^2+1

    So the induction hypothesis is:

    S_n=\sum^{n}_{r=1}(r^2+1)r!=n(n+1)!

    \Rightarrow S_{n+1} = S_n+\Big((n+1)^2+1\Big)(n+1)!

    =n(n+1)! + (n^2+2n+2)(n+1)!

    =(n+1)!(n^2+3n+2)

    =(n+1)!(n+2)(n+1)

    =(n+2)!(n+1)

    ...etc

    Grandad
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