Algebraic Manipulation What is the best method to get the LHS equal to RHS? 
$\frac{n(n+1)(n+2)}{3} + (n+1)(n+2) = \frac{(n+1)(n+2)(n+3)}{3}$
Thank you.

Firstly, take $(n+1)(n+2)$ out as a factor:
$((n+1)(n+2))[\frac n3 +1] = \frac{n+3}{3} (n+1)(n+2)$
Then divide through by $(n+1)(n+2)$:
$(\frac n3 +1) = \frac{n+3}{3}$
Multiply both sides by $3$:
$n+3 = n+3$
Therefore LHS = RHS
