Series and Sequences, Sigma notation.

**nicjoelim** · June 19th 2012, 10:26 AM

Given that r^2 +1 = (r+2)(r+1) - 3(r+1) + 2

Show that Sigma r=1 r=n (r^2 + 1)r! = n(n+1)!

Help please ><

**richard1234** · June 19th 2012, 10:46 AM

I understand that you're new here, welcome to MHF. Next time, you may want to wrap [tex] tags, so your question will look like this:

Given that $r^2 + 1 = (r+2)(r+1) - 3(r+1) + 2$

Show that $\sum_{r=1}^{r=n} (r^2 + 1)r! = n(n+1)!$ .

Anyway, replace $r^2 + 1$ with $(r+2)(r+1) - 3(r+1) + 2$ so that our summation becomes

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

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

$=(3! + ... + (n+2)!) - 3(2! + ... + (n+1)!) + 2(1! + ... + n!)$

Note that if $3 \le i \le n$ , all of the $i!$ terms cancel out, leaving

$(n+1)! + (n+2)! - 3(2! + (n+1)!) + 2(1! + 2!)$

$=(n+2)! - 2(n+1)!$

$=(n+2)(n+1)! - 2(n+1)!$ , this factors to $n(n+1)!$ .

**nicjoelim** · June 19th 2012, 10:51 AM

OMG Thank You

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