# Method of difference (proving)

• Sep 6th 2011, 06:12 AM
Punch
Method of difference (proving)
Making use of the method of differences, prove that $\sum_{n = 1}^{k}\frac{(n+2)(n+1)+3(n+2)-4}{(n+2)!}=3-\frac{1}{(k+1)!}-\frac{4}{(k+2)!}$

$\sum_{n=1}^{k}\frac{(n+2)(n+1)+3(n+2)-4}{(n+2)!}=\sum_{n = 1}^{k}\frac{(n+2)(n+1)}{(n+2)!}+\frac{3(n+2)}{(n+2 )!}-\frac{4}{(n+2)!}$

$=\sum_{n = 1}^{k}\frac{(n+2)(n+1)}{(n+2)(n+1)n!}+\frac{3(n+2) }{(n+2)(n+1)!}-\frac{4}{(n+2)!}$

$=\sum_{n = 1}^{k}\frac{1}{n!}+\frac{3}{(n+1)!}-\frac{4}{n+2)!}$
• Sep 15th 2011, 09:59 PM
CaptainBlack
Re: Method of difference (proving)
Quote:

Originally Posted by Punch
Making use of the method of differences, prove that $\sum_{n = 1}^{k}\frac{(n+2)(n+1)+3(n+2)-4}{(n+2)!}=3-\frac{1}{(k+1)!}-\frac{4}{(k+2)!}$

$\sum_{n=1}^{k}\frac{(n+2)(n+1)+3(n+2)-4}{(n+2)!}=\sum_{n = 1}^{k}\frac{(n+2)(n+1)}{(n+2)!}+\frac{3(n+2)}{(n+2 )!}-\frac{4}{(n+2)!}$

$=\sum_{n = 1}^{k}\frac{(n+2)(n+1)}{(n+2)(n+1)n!}+\frac{3(n+2) }{(n+2)(n+1)!}-\frac{4}{(n+2)!}$

$=\sum_{n = 1}^{k}\frac{1}{n!}+\frac{3}{(n+1)!}-\frac{4}{n+2)!}$

If you write out the first few terms you will find that the only terms at the begining of the summation that do not cancel out are:

$1+\frac{3}{2!}+\frac{1}{2!}=3$

similarly if you look at the middle of the summation you will see that all terms cancel and at the end the terms that do not cancel are:

$- \frac{1}{(k+1)!}-\frac{4}{(k+2)!}$

Alternatively for k even you can turn this into a more obviously telescoping series by combining terms in pairs.

CB