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

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 $\displaystyle {(n+1)(n+2)(n+3)(n+4)}/{4}$