Induction

**jennifer1004** · January 5th 2009, 11:34 AM

$P(n)=1*2*3+2*3*4+..+n(n+1)(n+2)={n(n+1)(n+2)(n+3)}/{4}$

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

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

Now I can't figure out how to get my answer to equal ${(n+1)(n+2)(n+3)(n+4)}/{4}$

**skeeter** · January 5th 2009, 11:42 AM

$P(n+1) = P(n) + (n+1)(n+2)(n+3)$

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

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

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

**jennifer1004** · January 5th 2009, 11:48 AM

How did you get from

to

**Danny** · January 5th 2009, 11:54 AM

Originally Posted by jennifer1004

How did you get from

to

Factor like terms.

**jennifer1004** · January 5th 2009, 12:00 PM

Duh! I feel silly. Thanks a bunch!

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Induction

How did you simplify that?

Thanks

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