# Induction

• Jan 5th 2009, 11:34 AM
jennifer1004
Induction
$\displaystyle P(n)=1*2*3+2*3*4+..+n(n+1)(n+2)={n(n+1)(n+2)(n+3)}/{4}$

Assume P(n), then P(n+1)=$\displaystyle 1*2*3+2*3*4+..+n(n+1)(n+2)+(n+1)(n+2)(n+3)=$ $\displaystyle {(n+1)(n+2)(n+3)(n+4)}/{4}$

So, P(n+1) = $\displaystyle 1*2*3+2*3*4+..+n(n+1)(n+2)+(n+1)(n+2)(n+3)$
= $\displaystyle {n(n+1)(n+2)(n+3)}/{4}+(n+1)(n+2)(n+3)$
= $\displaystyle (k+1)(k+2)(k+3)({k}/{4}+1)$

Now I can't figure out how to get my answer to equal $\displaystyle {(n+1)(n+2)(n+3)(n+4)}/{4}$
• Jan 5th 2009, 11:42 AM
skeeter
$\displaystyle P(n+1) = P(n) + (n+1)(n+2)(n+3)$

$\displaystyle P(n+1) = \frac{n(n+1)(n+2)(n+3)}{4} + \frac{4(n+1)(n+2)(n+3)}{4}$

$\displaystyle P(n+1) = \frac{n(n+1)(n+2)(n+3) + 4(n+1)(n+2)(n+3)}{4}$

$\displaystyle P(n+1) = \frac{(n+1)(n+2)(n+3)(n + 4)}{4}$
• Jan 5th 2009, 11:48 AM
jennifer1004
How did you simplify that?
• Jan 5th 2009, 11:54 AM
Jester
Quote:

Originally Posted by jennifer1004

Factor like terms.
• Jan 5th 2009, 12:00 PM
jennifer1004
Thanks
Duh! I feel silly. Thanks a bunch!