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Math Help - Prove

  1. #1
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    Prove

    This is a example question on a practice exam, any help would be appreciated. It is attached. Using induction to prove it.
    Thanks
    Attached Thumbnails Attached Thumbnails Prove-example4.jpg  
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  2. #2
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    Can't access the attachment
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  3. #3
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    For all integer n \geq 1, \sum_{i=1}^{n} i \cdot i! = (n+1)! - 1

    Sounds like your typical induction problem. Skipping the formalities ...

    Inductive step
    Assume it is true for n = k, i.e. \sum_{i=1}^{k} i \cdot i! = (k+1)! - 1.
    It remains to show that it is also true for n = k+1, i.e. \sum_{i=1}^{k+1} i \cdot i! = (k+2)! - 1 \qquad {\color{blue}(1)}

    But:
    \begin{aligned} \sum_{i=1}^{k+1} i \cdot i! & = \overbrace{{\color{red}1 \cdot 1! + 2\cdot 2! + \cdots + k \cdot k!}}^{\displaystyle \text{This is: } \sum_{i=1}^{k} i \cdot i!} + (k+1) \cdot (k+1)! \\ & = {\color{red}(k+1)! - 1} + (k+1)\cdot(k+1)! \qquad {\color{red} \text{By assumption}}\end{aligned}

    Factor out (k+1)! from the expression and you should get it equal to {\color{blue}(1)}
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