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 ><

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- Jun 19th 2012, 10:26 AMnicjoelimSeries and Sequences, Sigma notation.
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 >< - Jun 19th 2012, 10:46 AMrichard1234Re: Series and Sequences, Sigma notation.
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 $\displaystyle r^2 + 1 = (r+2)(r+1) - 3(r+1) + 2$

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

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

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

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

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

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

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

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

$\displaystyle =(n+2)(n+1)! - 2(n+1)!$, this factors to $\displaystyle n(n+1)!$. - Jun 19th 2012, 10:51 AMnicjoelimRe: Series and Sequences, Sigma notation.
OMG Thank You :)