# Summation proving 2

• May 20th 2011, 08:45 PM
Punch
Summation proving 2
Making use of the method of differences, prove that $\sum_{n = 1}^{k}\frac{n^2+6n+4}{(n+2)!}=3-\frac{1}{(k+1)!}-\frac{4}{(k+2)!}$

$\sum_{n = 1}^{k}\frac{n^2+6n+4}{(n+2)!}=\sum_{n=1}^{k}\frac{ (n+2)(n+1)+3(n+2)-4}{(n+2)!}$

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

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

$=\sum_{n = 1}^{k}\frac{1}{n!}+\frac{3}{(n+1)!}-\frac{4}{n+2)!}$
• May 20th 2011, 08:51 PM
TheChaz
Your first equality is false.
• May 20th 2011, 11:39 PM
Punch
I missed out something, the first part of the question asked to verify that $n^2+6n+4=(n+2)(n+1)+3(n+2)-4$, so it should be true
• May 20th 2011, 11:43 PM
TheChaz
Quote:

Originally Posted by Punch
I missed out something, the first part of the question asked to verify that $n^2+6n+4=(n+2)(n+1)+3(n+2)-4$, so it should be true

I was referring to the denominator.
• May 21st 2011, 05:39 AM
Punch
Quote:

Originally Posted by TheChaz
I was referring to the denominator.

sorry it was a typo, i corrected it