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Math Help - 1(1!)+ 2(2!) + 3(3!) + ... + n (n!)

  1. #1
    rcs
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    1(1!)+ 2(2!) + 3(3!) + ... + n (n!)

    i tried to make a move but i couldn't think where to start please help

    Find the sum: 1(1!)+ 2(2!) + 3(3!) + ... + n (n!)
    Define: n! = n(n-1) (n-2)(n-3) ... (3)(2)(1)

    thanks
    Last edited by mr fantastic; September 20th 2010 at 09:10 PM. Reason: Re-titled.
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  2. #2
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    Quote Originally Posted by rcs View Post
    i tried to make a move but i couldn't think where to start please help

    Find the sum: 1(1!)+ 2(2!) + 3(3!) + ... + n (n!)
    Define: n! = n(n-1) (n-2)(n-3) ... (3)(2)(1)

    thanks
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  3. #3
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    Quote Originally Posted by rcs View Post
    i tried to make a move but i couldn't think where to start please help

    Find the sum: 1(1!)+ 2(2!) + 3(3!) + ... + n (n!)
    Define: n! = n(n-1) (n-2)(n-3) ... (3)(2)(1)

    thanks
    Clearly:

    1! = 2! - 1.

    1(1!)+ 2(2!) = 3! - 1.

    1(1!)+ 2(2!) + 3(3!) = 4! -1.

    There is a clear pattern that suggests a hypothesis for the general formula. It is not difficult to prove the hypothesis true using induction.
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  4. #4
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    proof without induction: since k \cdot k! = (k+1)! - k!, we'll have a telescoping sum:

    \sum_{k=1}^n k \cdot k! = \sum_{k=1}^n [(k+1)! - k!]=(n+1)!-1!=(n+1)!-1.
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