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!
 A: HINT
The best way is what indicated by Raptors in the comments that is
$$\Sigma_{i=1}^n i^2=\frac {n(n+1)(2n+1)}6 \implies \Sigma_{i=1}^n (2i)^2=4\Sigma_{i=1}^n i^2=\frac {2n(2n+1)(2n+2)}6$$
then it suffices to prove by induction the LHS as explained in many duplicates you can find on the site as for example here


*

*Sum of First $n$ Squares Equals $\frac{n(n+1)(2n+1)}{6}$
For a general insight about the method we are applying refer to


*

*proof by induction - explanation on it
A: Here is a proof by weak induction.
The idea is to prove the statement for the smallest natural number possible, and then prove it for every number after that. Imagine climbing a ladder. You start at the bottom step and you have to walk up every step after that to get to the top. 
Let $P(n)$ be $$\sum_{i = 1}^{n} (2i)^2 = \frac{2n(2n+1)(2n+2)}{6}$$
Step one is the Base Case: Choose n = 1 because the summation formula starts at i = 1, so it doesn't make sense to start at 0 or a larger natural number. 
I. BASE CASE, $P(1):$ $$\sum_{i = 1}^{1} (2i)^2 = 2(1)^2 = \frac{2(1)(2(1)+1)(2(1)+2)}{6} = 2$$
Next is the assumption step, choose a variable that hasn't been used for clarity. 
II. Assume $P(j):$ $$\sum_{i = 1}^{j} (2i)^2 = \frac{2j(2j+1)(2j+2)}{6}$$
Now it must be shown that the statement is true for all following natural numbers. To help guide you write out $P(j+1)$. THIS DOES NOT PROVE ANYTHING. It is just a roadmap to help guide you through the problem.
$(*)$ $$\sum_{i = 1}^{j+1} (2i)^2 = \frac{2(j+1)(2(j+1)+1)(2(j+1)+2)}{6}$$
Here is the actual final step:
III. Show $P(j+1):$
$$\sum_{i = 1}^{j+1} (2i)^2 = [\sum_{i = 1}^{j} (2i)^2 ] + 4(j+1)^2$$
The following line is where we use the inductive hypothesis. What that means is that we use the assumption $P(j)$ to prove $P(j+1)$. When equal signs are involved this means making a substitution. 
$$\sum_{i = 1}^{j+1} (2i)^2 = [\frac{2j(2j+1)(2j+2)}{6} ] + 4j^2+8j+4$$
$$\sum_{i = 1}^{j+1} (2i)^2 = [\frac{(8j^3+12j^2+4j)}{6} ] + \frac{24j^2+48j+24}{6}$$
$$\sum_{i = 1}^{j+1} (2i)^2 = \frac{8j^3+36j^2+52j+24}{6}$$
At this point you can use synthetic division to factor the right hand side, but I'll leave that to you.
$$\sum_{i = 1}^{j+1} (2i)^2 = \frac{2(j+1)(2(j+1)+1)(2(j+1)+2)}{6}$$
You then end the proof by weak induction with the following statement:
$P(1)$ and $P(k+1)$ are true, therefore $\forall n\in \mathbb{N}($$\sum_{i = 1}^{n} (2i)^2 = \frac{2n(2n+1)(2n+2)}{6})$
A: I'm not sure where you got
$$
\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.
A: 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.
