Prove that $\sum^n_{k=1} k^2 = \binom{n+1}{2} + 2\binom{n+1}{3}$ for $n\geq 2$ Prove, for all $n\geq 2$ that 
$$\sum^n_{k=1} k^2 = \binom{n+1}{2} + 2\binom{n+1}{3}$$
Let us prove the inductive base for $n = 2$
$$\rm{LHS} = 1^2  + 2^2= 1 + 4 = 5$$
$$\rm{RHS} = \binom{3}{2} + 2\binom{3}{3} = 3 + 2\cdot 1 = 5$$
$$\rm{LHS} = \rm{RHS}$$
as desired.
Now, assume for some $k$, 
$$1^2 + 2^2 + \dots + k^2 = \binom{k+1}{2} + 2\binom{k+1}{3}$$
To prove the inductive step, it is sufficient to prove that, 
$$\binom{k+2}{2} + 2\binom{k+2}{3} - \binom{k+1}{2} - 2\binom{k+1}{3} = (k+1)^2$$
The LHS can be simplified as:
$$\frac{(k+2)!}{2!k!} + 2\left[\frac{(k+2)!}{3!(k-1)!}\right] - \frac{(k+1)!}{2!(k-1)!} - 2\left[\frac{(k+1)!}{3!(k-2)!}\right]$$
$$=\frac{3(k+2)! + 2k(k+2)! -3k(k-1)! - 2k(k-1)(k+1)!}{6k!}$$
$$=\frac{(k+1)!(3k + 6 - 2k^2 - 2k) + (k-1)!(2k^3 + 2k^2 - 3k)}{6k!}$$
$$=\frac{(k+1)!(-2k^2+k+6) + k!(2k^2+2k - 3)}{6k!}$$
$$=\frac{(k+1)(-2k^2 + k + 6) + 2k^2 +2k - 3}{6}$$
$$=\frac{-2k^3 + k^2 + 9k + 3}{6}$$
I'm pretty sure I'm making a mistake somewhere, but I can't figure it out. If someone could complete this inductive proof for me, I will be grateful.
Also, I don't feel satisfied with this ugly proof. Please add a combinatorial proof for this in your answer if possible. 

Perhaps this Wikipedia link might prove itself useful. The sums are computed through telescoping, each relying  on the result of the previous one, which, in this case, is $\sum_1^n k=\frac{n(n+1)}2$, for instance like this.
