proving this summation

• Sep 30th 2009, 04:04 AM
thereddevils
proving this summation
Prove that $\sum^{n}_{r=1}[(r+2)(r+1)-3(r+1)+2]r!=n(n+1)!$

Thanks a lot .
• Sep 30th 2009, 05:36 AM
Amer
Quote:

Originally Posted by thereddevils
Prove that $\sum^{n}_{r=1}[(r+2)(r+1)-3(r+1)+2]r!=n(n+1)!$

Thanks a lot .

let prove it by induction okay

let $P_n : \sum_{r=1}^{n} [(r+2)(r+1) - 3(r+1) +2]r! = n(n+1)!$

first see if n=1 and 2 true

$(1+2)(1+1)-3(2)+2 = 2$ and

$(1)(2)! = 2$ at 1 the statement is true try when n=2

you will find it is true now suppose that n is true try n+1

$P_{n+1}:
\sum_{r=1}^{n+1} [(r+2)(r+1)-3(r+1)+2]r! = (n+1)(n+2)!$
......(*)

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

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

$=(n+1)![n+(n+3)(n+2)-3(n+2)+2]$

$=(n+1)![n+n^2+5n+6-3n-6+2]$

$=(n+1)!(n^2+3n+2)$

$=(n+1)!(n+1)(n+2)=(n+1)(n+2)!$ you begin with the RHS in (*) and finished with LHS in the (*) formula

so P_{n+1} is true the proof end
• Sep 30th 2009, 07:38 AM
Hello thereddevils
Quote:

Originally Posted by thereddevils
Prove that $\sum^{n}_{r=1}[(r+2)(r+1)-3(r+1)+2]r!=n(n+1)!$

Thanks a lot .

It's much easier if you simplify the expression first!

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

So the induction hypothesis is:

$S_n=\sum^{n}_{r=1}(r^2+1)r!=n(n+1)!$

$\Rightarrow S_{n+1} = S_n+\Big((n+1)^2+1\Big)(n+1)!$

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

$=(n+1)!(n^2+3n+2)$

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

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

...etc