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!

I'm not sure where you got
$$
\frac{2n(2n+1)(2n+2)(12(n+1)^2)}{6}
$$
from. It should be
$$
\frac{2n(2n+1)(2n+2)\color{red}{+}12(n+1)^2}{6}
$$
and with this fix you should be able to finish up.

However, it's better if you do some simplifications, to get rid of useless factors.
The induction step consists in writing
$$
\sum_{i=1}^{n+1}(2i)^2=
\sum_{i=1}^{n}(2i)^2+(2(n+1))^2
$$
and substituting the summation with the formula provided by the induction hypothesis, so to get
$$
\frac{2n(2n+1)(2n+2)}{6}+4(n+1)^2
$$
Your goal is to prove that this equals
$$
\frac{2(n+1)(2(n+1)+1)(2(n+1)+2)}{6}=
\frac{2(n+1)(2n+3)(n+2)}{3}
$$
which you obtain from the formula to prove by changing $n$ into $n+1$, which you seem to have done right.
Consider
\begin{align}
\frac{2n(2n+1)(2n+2)}{6}+4(n+1)^2
&=\frac{2n(2n+1)(n+1)}{3}+4(n+1)^2 \\[4px]
&=\frac{2}{3}(n+1)\bigl(n(2n+1)+6(n+1)\bigr) \\[4px]
&=\frac{2}{3}(n+1)(2n^2+7n+6)
\end{align}
Since $2n^2+7n+6=(n+2)(2n+3)$, you're done.
