Hi all,
Prove by induction that $\displaystyle \sum_{r=1}^n r(r!)$= (n+1)!-1
for all integers n>=1
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.