Prove using induction that $1^2 + 3^2 + 5^2 + . . . + (2n + 1)^2 = (n + 1) ∗ (2n + 1) ∗ (2n + 3)/3$ My question reads as follows: 
Prove using induction that
$$1^2 + 3^2 + 5^2 + . . . + (2n + 1)^2 = (n + 1) \times (2n + 1) \times (2n + 3)/3.$$
Firstly, I must prove the base case. For $n=0$, my LHS $= 1$ and my RHS $= 1$. Perfect, I've proven the base case.
I must now assume that $n=k$ for some arbitrary $ k\geq 0$. 
Secondly, I must assume the induction hypothesis which states that
$S(k) = (k+1)(2k+1)(2k+3)/3$. 
We want to prove that for some arbitrary $k$, the predicate of $k+1$ also holds true.
I get to this point and everything is fine until I am unable to return an answer that matches my presumed $S(k+1)$. My answer must be done in this format in order to get full marks.
So far what I have is the following:
In order to get the $(k+1)$th term, we must add the $(k+1)$-th term to the series. Therefore it looks like the following:
$$1^2 + 3^2 + 5^2 + ... + (2k+1)^2 + \mathbf{(2k+3)^2}$$
By substituting in $S(k)$ we get the following:
$$(k+1)(2k+1)(2k+3)/3 + (2k+3)^2$$
It is here that I encounter an issue. No matter which simplification techniques I use, after this step I can never return an answer that is $(k+2)(2k+3)(2k+5)/3$ so I am stuck. And unfortunately, I am not allowed to then assume we can rewrite the series in terms of $(2k-1)^2$ which would make things too easy and I could easily get back the right-hand equivalent to that. In desperate need of help. Thanks.
 A: $\begin{align}\frac 13 (k+1)(2k+1)(2k+3)+(2k+3)^2&=\frac 13(2k+3)\bigg((k+1)(2k+1)+3(2k+3)\bigg)
\\&=\frac 13(2k+3)\bigg(2k^2+3k+1+6k+9\bigg)
\\&=\frac 13(2k+3)\bigg(2k^2+9k+10\bigg)
\\&=\frac 13(2k+3)(k+2)(2k+5)
\end{align}$
A: So we want to show that when $n=k+1$ we have
$1^2+3^2+\cdot\cdot\cdot+(2k+1)^2+(2k+3)^2 = \frac{(k+2)(2k+3)(2k+5)}{3}$.
So what I like to do is some scratch work, we'll take 
$(k+1)(2k+1)(2k+3)(3^{-1})+(2k+3)^2$
and expand it out, (I'll leave this to you). Then after doing that you'll also want to expand $\frac{(k+2)(2k+3)(2k+5)}{3}$. What you should notice is that you get the same thing. 
So when you write up your solution just expand,
$(k+1)(2k+1)(2k+3)(3^{-1})+(2k+3)^2$,
then simplify by reverse your steps from when you expanded $\frac{(k+2)(2k+3)(2k+5)}{3}$. 
A: $\frac {(k+1)(2k+1)(2k+3)}3 + (2k+3)^2 = \frac {(k+2)(2(k+1)+1)(2(k+1)+3)}3 \iff$
$\frac {(k+1)(2k+1)(2k+3)+ 3(2k+3)^2}3= \frac {(k+2)(2(k+1)+1)(2(k+1)+3)}3 \iff$
$(k+1)(2k+1)(2k+3)+ 3(2k+3)^2 = (k+2)(2(k+1)+1)(2(k+1)+3)\iff$
$(2k+3)[(k+1)(2k+1) + 3(2k+3)] = (k+2)(2k+3)(2k+5)\iff$
$[(k+1)(2k+1) + 3(2k+3)]=(k+2)(2k+5)$ or $2k+3=0$ which as $k$ is natural never happens$\iff$
$[(2k^2 + 3k + 1) + (6k + 9)= 2k^2 +9k + 10\iff$
$3k^2 + 9k +10 = 2k^3 + 9k + 10$
Which it does
..... 
Or just do it:
$\frac {(k+1)(2k+1)(2k+3)}3 + (2k+3)^2=$
$\frac {(k+1)(2k+1)(2k+3)+ 3(2k+3)^2}3=$
$\frac{(2k^2 + 3k + 1)(2k+3) +3(4k^2 + 12k+9)}3=$
$\frac {4k^3 + 24k^2 + 47k + 30}3$.
While $\frac {(k+2)(2(k+1)+1)(2(k+1)+3)}3 =$
$\frac {(k+2)(2k+3)(2k+5)}3 =$
$\frac {(2k^2 + 7k + 6)(2k+5)}3=$
$\frac {4k^3 + 24k^2 + 47k + 30}3$.
Those are both equal.
