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Math Help - Ugh, another inductive proof

  1. #1
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    Ugh, another inductive proof

    I have to prove this inductively, bare with me if the lingo isn't perfect.

    Prove that: 1(1!) + 2(2!) + ... + n(n!) = (n+1)! -1

    Proof:
    Base Case: n=1;
    LHS = 1(1!) = 1
    RHS = (1+1)! - 1 = 2! - 1 = 1
    Done;

    Now, assume true for n; Prove true for n=n+1.
    Let n= n+1
    RHS = ( (n+1)+1)! - 1 = (n + 2)! - 1
    LHS = 1(1!) + 2(2!) + ... + n(n!) + (n+1)((n+1)!)

    now, since we stated that 1(1!) + 2(2!) + ... + n(n!) = (n+1)! -1 we can sub that in its place such that the equation looks like the following:

    LHS = (n+1)! - 1 + (n+1)((n+1)!)

    Now here is where I am stuck ... I have tried factoring and combining like terms in every way I can think of ... anyone want to lend a hand =D
    TY
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  2. #2
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    Quote Originally Posted by mcox05 View Post
    Prove this inductively 1(1!) + 2(2!) + ... + n(n!) = (n+1)! -1
    You have the base case.
    Suppose that this is true: \sum\limits_{k = 1}^N {k\left( {k!} \right)}  = \left( {N + 1} \right)! - 1.
    Note that: \left( {N + 2} \right)! - 1 = \left( {N + 1} \right)!\left[ {N + 2} \right] - 1 = \left( {N + 1} \right)!\left[ {1 + \left( {N + 1} \right)} \right] - 1
    \begin{array}{lcl}<br />
   {\sum\limits_{k = 1}^{N + 1} {k\left( {k!} \right)} } &  =  & {\sum\limits_{k = 1}^N {k\left( {k!} \right)}  + \left( {N + 1} \right)\left[ {\left( {N + 1} \right)!} \right]}  \\<br />
   {} &  =  & {\left[ {\left( {N + 1} \right)! - 1} \right] + \left( {N + 1} \right)\left[ {\left( {N + 1} \right)!} \right]}  \\<br />
   {} &  =  & {\left( {N + 2} \right)! - 1}  \\<br />
 \end{array}
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