# Thread: Proof by induction

1. ## Proof by induction

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

2. 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 $n \geq 1$ .