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Math Help - [SOLVED] Need some help with an induction proof

  1. #1
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    [SOLVED] Need some help with an induction proof

    I'm having some issues with the inductive step for this problem, and I could use a hand with it.

    Let F(n) be the statement that 1\cdot1!+2\cdot2!+...+n\cdot n!=(n+1)!-1 whenever n is a positive integer.

    I have the basis step and inductive hypothesis done, but I'm having some trouble with the induction proof for F(k+1). Does anyone have any advice?
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  2. #2
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    Quote Originally Posted by Runty View Post
    I'm having some issues with the inductive step for this problem, and I could use a hand with it.

    Let F(n) be the statement that 1\cdot1!+2\cdot2!+...+n\cdot n!=(n+1)!-1 whenever n is a positive integer.

    I have the basis step and inductive hypothesis done, but I'm having some trouble with the induction proof for F(k+1). Does anyone have any advice?
    Hi Runty,

    Sure!

    F(n)

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

    If this is so, then the following must be true because of that

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

    To discover whether or not this is so, we utilise statement (1)

    F(k+1)

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

    =(k+1)!(1+k+1)-1

    =(k+1)!(k+2)-1=(k+2)!-1
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  3. #3
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    Quote Originally Posted by Runty View Post
    I'm having some issues with the inductive step for this problem, and I could use a hand with it.

    Let F(n) be the statement that 1\cdot1!+2\cdot2!+...+n\cdot n!=(n+1)!-1 whenever n is a positive integer.
    Say that we know that \sum\limits_{k = 1}^N {k \cdot k!}  = \left( {N + 1} \right)! - 1
    Then
    <br />
\begin{array}{rcl}<br />
   {\sum\limits_{k = 1}^{N + 1} {k \cdot k!} } &  =  & {\sum\limits_{k = 1}^N {k \cdot k!}  + \left( {N + 1} \right) \cdot \left( {N + 1} \right)!}  \\<br />
   {} &  =  & {\left( {N + 1} \right)! - 1 + \left( {N + 1} \right) \cdot \left( {N + 1} \right)!}  \\<br />
   {} &  =  & {\left( {N + 1} \right)! \cdot (N + 2) - 1}  \\<br /> <br />
 \end{array}
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