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Math Help - Help proving this identity?

  1. #1
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    Help proving this identity?

    Hey guys, im looking for step by step help proving this identity. Im sans a calc book right now and i cant figure it out...

    Again, a step by step would be most useful, my calculus is a little rusty, and i'm having to do this stuff for a computer algorithms course. I dont just need this particular answer, i need to be able to reproduce it on other questions.

    Thanks a ton!

    <br /> <br />
\sum^{n}_{k=1} k(k!) = (n + 1)! - 1<br /> <br />
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  2. #2
    MHF Contributor alexmahone's Avatar
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    Quote Originally Posted by Dfowj View Post
    Hey guys, im looking for step by step help proving this identity. Im sans a calc book right now and i cant figure it out...

    Again, a step by step would be most useful, my calculus is a little rusty, and i'm having to do this stuff for a computer algorithms course. I dont just need this particular answer, i need to be able to reproduce it on other questions.

    Thanks a ton!

    <br /> <br />
\sum^{n}_{k=1} k(k!) = (n + 1)! - 1<br /> <br />
    This sum telescopes.

    \sum^{n}_{k=1} k(k!) =\sum^{n}_{k=1} (k+1-1)(k!)

    =\sum^{n}_{k=1} (k+1)(k!) - k!

    =\sum^{n}_{k=1} (k+1)! - k!

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

    =(n+1)!-1
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  3. #3
    o_O
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    Try induction. For the inductive step, you assume that it is true for n and you want to show that this implies this is true for n+1, i.e. we want to show that \sum_{k=1}^{n+1} k \cdot k! = \left(n+2\right)! - 1

    So start with the LHS and notice that:
    \begin{aligned} \sum_{k=1}^{n+1} k \cdot k! & = {\color{red}\sum_{k=1}^n k \cdot k!} + (n+1)(n+1)! \\ & = {\color{red}(n+1)!} + (n+1)(n+1)! {\color{red} \ - 1} \qquad (\text{By our } {\color{red} \text{assumption}}) \\ &  \ \vdots \end{aligned}

    Factor and you'll get what we were after. So by the principle of mathematical induction, your identity is true for all n \geq 1
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