Thanks for taking my question:
How can I prove by induction the following:
summation E (i=1 to n) of i * i! = (n+1)! -1
- thanks
no option
no option again
Hi,
is it $\displaystyle \sum_{i=1}^n i\cdot i! = (n+1)!-1$ ?
base case: $\displaystyle n=1$ ... $\displaystyle 1\cdot 1!=(1+1)!-1$ verified.
inductive step: suppose $\displaystyle \sum_{i=1}^n i\cdot i! = (n+1)!-1$ and let's prove it holds for $\displaystyle n+1$ too, i.e. we want to prove $\displaystyle \sum_{i=1}^{n+1} i\cdot i!=((n+1)+1)!-1$.
$\displaystyle \sum_{i=1}^{n+1} i\cdot i! = \sum_{i=1}^n i\cdot i! + (n+1)\cdot (n+1)!=(n+1)!-1+(n+1)\cdot (n+1)! = $
$\displaystyle =(n+1)!(1+(n+1))-1=(n+1)!(n+2)-1=(n+2)!-1=((n+1)+1)!-1$ inductive step verified.