Prove that $\displaystyle \sum^{n}_{r=1}[(r+2)(r+1)-3(r+1)+2]r!=n(n+1)!$
Thanks a lot .
let prove it by induction okay
let $\displaystyle P_n : \sum_{r=1}^{n} [(r+2)(r+1) - 3(r+1) +2]r! = n(n+1)!$
first see if n=1 and 2 true
$\displaystyle (1+2)(1+1)-3(2)+2 = 2 $ and
$\displaystyle (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
$\displaystyle P_{n+1}:
\sum_{r=1}^{n+1} [(r+2)(r+1)-3(r+1)+2]r! = (n+1)(n+2)!$ ......(*)
$\displaystyle \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)!}+ $$\displaystyle [(n+3)(n+2)-3(n+2)+2](n+1)!$
$\displaystyle \sum_{r=1}^{n+1}[(r+2)(r+1)-3(r+1)+2]r! = n(n+1)!+[(n+3)(n+2)-3(n+2)+2](n+1)!$
$\displaystyle =(n+1)![n+(n+3)(n+2)-3(n+2)+2]$
$\displaystyle =(n+1)![n+n^2+5n+6-3n-6+2]$
$\displaystyle =(n+1)!(n^2+3n+2)$
$\displaystyle =(n+1)!(n+1)(n+2)=(n+1)(n+2)!$ you begin with the RHS in (*) and finished with LHS in the (*) formula
so P_{n+1} is true the proof end
Hello thereddevilsIt's much easier if you simplify the expression first!
$\displaystyle (r+2)(r+1)-3(r+1)+2 = r^2+1$
So the induction hypothesis is:
$\displaystyle S_n=\sum^{n}_{r=1}(r^2+1)r!=n(n+1)!$
$\displaystyle \Rightarrow S_{n+1} = S_n+\Big((n+1)^2+1\Big)(n+1)!$
$\displaystyle =n(n+1)! + (n^2+2n+2)(n+1)!$
$\displaystyle =(n+1)!(n^2+3n+2)$
$\displaystyle =(n+1)!(n+2)(n+1)$
$\displaystyle =(n+2)!(n+1)$
...etc
Grandad