<s>[INST] Proving that $1\cdot3+3\cdot5+5\cdot7+\cdots+(2n-1)(2n+1)={n(4n^2+6n-1) \over 3}$ by induction for $n\geq 1$ 
Prove using mathematical induction that
$$1\cdot3+3\cdot5+5\cdot7+\cdots+(2n-1)(2n+1)= {n(4n^2+6n-1) \over 3}.$$

Step 1: If we assume that the equation is true for a natural number, $n=k$, then we get
$$1\cdot3+3\cdot5+5\cdot7+\cdots+(2k-1)(2k+1)= {k(4k^2+6k-1) \over 3}$$
Step 2: When a statement is true for a natural number $n = k$,
then it will also be true for its successor, $n=k+1$. Hence, we have to prove that it is also true for $n=k+1$.
$$1\cdot3+3\cdot5+5\cdot7+\cdots+(2k-1)(2k+1)+(2k+1-1)(2k+1+1) = {k(4k+1^2+6k+1-1) \over 3}$$
I replace the LHS by step 1.
$${k(4k^2+6k-1) \over 3} + 2(k+1-1)(2k+1+1)={k(4k+1^2+6k+1-1) \over 3}$$
Now I need to make LHS equal to RHS.
 [/INST] For what it's worth, I think Mario's answer is definitely the most elegant, but I know you are trying to prove the relation by induction. Thus, I will outline a proof by induction. 
Proof. For $n\geq 1$, let $S(n)$ denote the statement
$$
S(n) : \sum_{i=1}^n(2i-1)(2i+1)=\frac{n(4n^2+6n-1)}{3}.
$$
Base step ($n=1$): $S(1)$ says that $(2-1)(2+1)=3=\frac{4+6-1}{3}$, and this is true.
Induction step ($S(k)\to S(k+1)$): Fix some $k\geq 1$ and assume that
$$
S(k) : \sum_{i=1}^k(2i-1)(2i+1)=\frac{k(4k^2+6k-1)}{3}
$$
holds. To be shown is that
$$
S(k+1) : \sum_{i=1}^{k+1}(2i-1)(2i+1)=\frac{(k+1)[4(k+1)^2+6(k+1)-1]}{3}
$$
follows. 

Note: Later in the proof, it may help to first observe the following:
$$
(k+1)[4(k+1)^2+6(k+1)-1]=(k+1)(4k^2+14k+9)=4k^3+18k^2+23k+9\tag{$\dagger$}
$$

Beginning with the left-hand side of $S(k+1)$,
\begin{align}
\sum_{i=1}^{k+1}(2i-1)(2i+1)&= \sum_{i=1}^k(2i-1)(2i+1)+\bigl(2(k+1)-1\bigr)\bigl(2(k+1)+1\bigr)\\[1em]
&= \frac{4(k^2+6k-1)}{3}+(2k+1)(2k+3)\tag{by $S(k)$}\\[1em]
&= \frac{k(4k^2+6k-1)+(12k^2+24k+9)}{3}\\[1em]
&= \frac{4k^3+18k^2+23k+9}{3}\tag{simplify}\\[1em]
&= \frac{(k+1)[4(k+1)^2+6(k+1)-1]}{3},\tag{by $(\dagger)$}
\end{align}
one arrives at the right-hand side of $S(k+1)$, completing the inductive proof. 
By mathematical induction, the statement $S(n)$ is true for all $n\geq 1$. $\blacksquare$
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