# Thread: quick fraction simplification help

1. ## quick fraction simplification help

Hey, just need some help with a fraction simplification problem.

how can i make [k(k+1)(k+2)]/3 + (k+1)(k+2) = (k+1)(k+2)(k/3 + 1)

and then take that step and have it equal [(k+1)(k+2)(k+3)]/3

i slightly understand the second to third step but not how to go from the first to the second (factor out a k but how?) thanks.

2. Originally Posted by Xender1
Hey, just need some help with a fraction simplification problem.

how can i make [k(k+1)(k+2)]/3 + (k+1)(k+2) = (k+1)(k+2)(k/3 + 1)

and then take that step and have it equal [(k+1)(k+2)(k+3)]/3

i slightly understand the second to third step but not how to go from the first to the second (factor out a k but how?) thanks.

$\frac{k(k+1)(k+2)}{3}+(k+1)(k+2)$

$=\frac{k(k+1)(k+2)+3(k+1)(k+2)}{3}$

$=\frac{(k+1)(k+2)(k+3)}{3}$

$=(k+1)(k+2)(\frac{k+3}{3})$

$
=(k+1)(k+2)(k/3 + 1)
$