Prove

**captainjapan** · November 25th 2008, 07:18 PM

This is a example question on a practice exam, any help would be appreciated. It is attached. Using induction to prove it.
Thanks

**dolphinlover** · November 25th 2008, 07:47 PM

Can't access the attachment

**o_O** · November 25th 2008, 09:02 PM

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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