Summation proving 2

**Punch** · May 20th 2011, 08:45 PM

Making use of the method of differences, prove that $\sum_{n = 1}^{k}\frac{n^2+6n+4}{(n+2)!}=3-\frac{1}{(k+1)!}-\frac{4}{(k+2)!}$

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

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

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

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

**TheChaz** · May 20th 2011, 08:51 PM

Your first equality is false.

**Punch** · May 20th 2011, 11:39 PM

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

**TheChaz** · May 20th 2011, 11:43 PM

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.

**Punch** · May 21st 2011, 05:39 AM

Originally Posted by TheChaz

I was referring to the denominator.

sorry it was a typo, i corrected it

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