Help proving this identity?

**Dfowj** · August 30th 2009, 02:26 PM

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!

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

**alexmahone** · August 30th 2009, 02:40 PM

Originally Posted by Dfowj

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!

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

This sum telescopes.

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

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

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

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

$=(n+1)!-1$

**o_O** · August 30th 2009, 02:40 PM

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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