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

    I have the basis step and inductive hypothesis done, but I'm having some trouble with the induction proof for $\displaystyle F(k+1)$. 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 $\displaystyle F(n)$ be the statement that $\displaystyle 1\cdot1!+2\cdot2!+...+n\cdot n!=(n+1)!-1$ whenever $\displaystyle n$ is a positive integer.

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

    Sure!

    F(n)

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

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

    $\displaystyle (1)1!+(2)2!+(3)3!+....(n)n!+(n+1)(n+1)!=(n+2)!-1$

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

    F(k+1)

    $\displaystyle (1)1!+(2)2!+....(k)k!+(k+1)(k+1)!=(k+1)!-1+(k+1)(k+1)!$

    $\displaystyle =(k+1)!(1+k+1)-1$

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

    \end{array} $
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