Hi all,

Prove by induction that $\displaystyle \sum_{r=1}^n r(r!)$= (n+1)!-1

for all integers n>=1

Printable View

- Nov 5th 2009, 09:10 AMsirellwoodInduction Proof
Hi all,

Prove by induction that $\displaystyle \sum_{r=1}^n r(r!)$= (n+1)!-1

for all integers n>=1 - Nov 5th 2009, 01:52 PMDeadstar
Inductive base case.

Let n=1. Then you have for LHS

$\displaystyle \sum_{r=1}^1 r(r!) = 1(1!) = 1$

And RHS

(1+1)! - 1 = 2! - 1 = 2-1 = 1. So hold for n=1.

Assume it holds for n, i.e $\displaystyle \sum_{r=1}^n r(r!) = (n+1)! - 1$

Try to prove it works for n+1

So for n+1 we want to get $\displaystyle \sum_{r=1}^{n+1} r(r!) = ((n+1)+1)! - 1 = (n+2)! - 1$.

$\displaystyle \sum_{r=1}^{n+1} r(r!) = \sum_{r=1}^{n} r(r!) + (n+1)*(n+1)! = (n+1)! - 1 + (n+1)*(n+1)!$.

For the terms involving n take out a common factor of (n+1)! to get...

$\displaystyle (1+(n+1))*(n+1)! - 1 = (n+2)*(n+1)! - 1 = (n+2)! - 1$. which is what we wanted. Hence by our inductive hypothesis the equality holds. - Nov 5th 2009, 02:35 PMsirellwood
Wow. Great answer, very clear. You probably won't beleive me but i was kind of halway there, i showed it held for n=1 but got confused half way through showing it holds for n+1!