Proof by induction

**N.A.G** · September 20th 2008, 03:31 AM

I have been asked to prove that (1*1!)+(2*2!)+(3*3!)+ ... + (n*n!) is always = to (n+1)!-1 for all positive intergers using proof by induction. If anyone could help me on this one i would be most thankfull

Thanks for your help N.A.G

I just realised i put this in the wrong thread, as its not high school maths. for that im sorry but i dont know how to move it so unless some one tells me how to or tell me how to delet it it will have to stay here lol soz

**Sean12345** · September 20th 2008, 05:25 AM

Hi N.A.G,

You have been asked to prove

$\sum^n_{k=1}{(k\cdot k!)}=(n+1)!-1~~\forall n \in \mathbb {Z}$ and $n \geq 1$

by induction.

Let's define the above statement as $P(n)$ . First we shall prove that $P(1)$ is true and that $P(n)\implies P(n+1)$ is true. We have

$P(1)=(1\cdot1!)=(1+1)!-1=1$

thus $P(1)$ is true. Now assume that $P(n)$ is indeed true and hence giving,

$\sum^n_{k=1}{(k\cdot k!)}+[(n+1)(n+1)!]=[(n+1)!-1]+[(n+1)(n+1)!]$

If we look at the Right hand side and take out a common factor of $(n+1)!$ this leads to

$\sum^{n+1}_{k=1}{(k\cdot k!)}=(n+1)![1+(n+1)]-1=(n+1)!(n+2)-1$

Note that $(n+1)!(n+2)=(n+2)!$ thus giving

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

Hence we have proven $P(n)\implies P(n+1)$ . We conclude that since we have proven that $P(1)$ is true and that $P(n)\implies P(n+1)$ is true then by induction we have proven that $P(n)$ is true $\forall n \in \mathbb {Z}$ and $n \geq 1$ .

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