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Math Help - Number Theory Proof

  1. #1
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    Number Theory Proof

    I need to prove the following Proposition

    Let N be a positive integer. Then:
    1*1! + 2*2! + 3*3! + ... + n *n! = (n+1)! - 1
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by algebrapro18 View Post
    I need to prove the following Proposition

    Let N be a positive integer. Then:
    1*1! + 2*2! + 3*3! + ... + n *n! = (n+1)! - 1
    \sum_{r=1}^n r \times r! =\sum_{r=1}^n (r+1)!-r!

    This is a telescoping series and the sum is:

    \sum_{r=1}^n r \times r! =\sum_{r=1}^n (r+1)!-r!=(n+1)!-1

    That is all but the (n+1)! from the last term and the -1 from the first cancel
    with a corresponding term from the next term in the series.

    RonL
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  3. #3
    Behold, the power of SARDINES!
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    By induction..

    Quote Originally Posted by algebrapro18 View Post
    I need to prove the following Proposition

    Let N be a positive integer. Then:
    1*1! + 2*2! + 3*3! + ... + n *n! = (n+1)! - 1

    base case is true... n=1

    assume n=k

    1 \cdot 1! +... k \cdot (k)! = (k+1)!-1

    show k+1

    1 \cdot 1! +... k \cdot (k)! + (k+1) \cdot (k+1)!

    by hypothesis..

    (k+1)!-1 + (k+1) \cdot (k+1)!

    grouping and factoring

    (k+1!) + (k+1) \cdot (k+1)! -1 = (k+1)![1+(k+1)]-1=(k+2)!-1
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  4. #4
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    Thank you so much, I thought proof by induction was the way to go.

    Is there also a way to do this by Combinatorial Proof because thats the section this problem was from.
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