I need to prove the following Proposition
Let N be a positive integer. Then:
1*1! + 2*2! + 3*3! + ... + n *n! = (n+1)! - 1
$\displaystyle \sum_{r=1}^n r \times r! =\sum_{r=1}^n (r+1)!-r!$
This is a telescoping series and the sum is:
$\displaystyle \sum_{r=1}^n r \times r! =\sum_{r=1}^n (r+1)!-r!=(n+1)!-1$
That is all but the (n+1)! from the last term and the -1 from the first cancel
with a corresponding term from the next term in the series.
RonL
base case is true... n=1
assume n=k
$\displaystyle 1 \cdot 1! +... k \cdot (k)! = (k+1)!-1$
show k+1
$\displaystyle 1 \cdot 1! +... k \cdot (k)! + (k+1) \cdot (k+1)!$
by hypothesis..
$\displaystyle (k+1)!-1 + (k+1) \cdot (k+1)!$
grouping and factoring
$\displaystyle (k+1!) + (k+1) \cdot (k+1)! -1 = (k+1)![1+(k+1)]-1=(k+2)!-1$