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

Printable View

- Mar 9th 2008, 12:59 PMalgebrapro18Number Theory Proof
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 - Mar 9th 2008, 01:05 PMCaptainBlack
$\displaystyle \sum_{r=1}^n r \times r! =\sum_{r=1}^n (r+1)!-r!$

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 - Mar 9th 2008, 01:20 PMTheEmptySetBy induction..

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

$\displaystyle (k+1!) + (k+1) \cdot (k+1)! -1 = (k+1)![1+(k+1)]-1=(k+2)!-1$ - Mar 9th 2008, 03:52 PMalgebrapro18
Thank you so much, I thought proof by induction was the way to go.

Is there also a way to do this by Combinatorial Proof because thats the section this problem was from.