I would really appreciate some help solving this induction problem!! 
$3.$ for all $n\ge1$, $\displaystyle\sum_{i=1}^n(2i)^2=\frac{2n(2n+1)(2n+2)}{6}$

I have
$$\frac{2n(2n+1)(2n+2)(12(n+1)^2)}6= \frac{2n(2n+1)2(n+1)12(n+1)(n+1)}6.$$
I think I need to do something with the $(n+1)$.
However, I'm not sure where to go from here. I know the end goal is:
$$\frac{2(n+1)(2(n+1)+1)(2(n+1)+2)}6.$$
Any help is appreciated!

So to prove this we must use induction and  since the LHS is a sum, we add $(2(n+1))^2$.
We do this because if $n$ and $n+1$ every natural number works as $n$ is arbitrary. Therefore, we only need to prove it works for $n+1$.
$$\sum_{i=1}^{n+1}(2i)^2=\frac{2n(2n+1)(2n+2)}{6}+(2(n+1))^2$$
This gives:
$$\sum_{i=1}^{n+1}(2i)^2=\frac{(4n^2+14n+12)(2n+2)}{6}$$
$4n^2+14n+12$ factors to give:
$$\sum_{i=1}^{n+1}(2i)^2=\frac{(2n+3)(2n+4)(2n+2)}{6}$$
Therefore:
$$\sum_{i=1}^{n+1}(2i)^2=\frac{(2(n+1)+2)(2(n+1)+1)(2(n+1))}{6}$$
All that remains is to check the "base-case" when $n=1$.
This will complete the proof.
