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.
 A: From the formula for the sum of the square of the first $n$ natural numbers we have
\begin{align*}
\sum_{k=1}^n(2k-1)(2k+1)&=\sum_{k=1}^n(4k^2-1)\\
&=4\sum_{k=1}^nk^2-\sum_{k=1}^n1\\
&=\frac{4n(n+1)(2n+1)}{6}-n\\
&=\frac{n\left[4(n+1)(2n+1)-6\right]}{6}\\
&=\frac{n(4n^2+6n-1)}{3}
\end{align*}
A: You have this line: $$1\cdot3+3\cdot5+5\cdot7+\cdots+(2k-1)(2k+1)+(2k+1-1)(2k+1+1) $$$${}= {k(4k+1^2+6k+1-1) \over 3}$$
That should be: $$1\cdot3+3\cdot5+5\cdot7+\cdots+(2k-1)(2k+1)+(2(k+1)-1)(2(k+1)+1)$$$${} = {k(4k^2+6k-1) \over 3}+(2(k+1)-1)(2(k+1)+1)$$
First of all, I'm just doing literally the same thing to both sides of an already-assumed-to-be-true equation. I'm adding $(2(k+1)-1)(2(k+1)+1)$.
Second, the reason I've added this is because it is what the next term would look like. Not $(2k+1-1)(2k+1+1)$, which just inserted some "${}+1$" in some places.
From here, you need to give a common denominator to the right side and see that it works out to be ${(k+1)(4(k+1)^2+6(k+1)-1) \over 3}$.
A: 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 $n\geq 1$. $\blacksquare$
A: The general idea is the following:
Step 1 Check that your identity holds true for $n=n_0 $ (e.g 1)
Step 2 Assume that your identity holds true for $n=k>n_0$ and save the result.
Step 3 Explore if the identity holds true for $k+1$ using the result in step 2.
If step 1 holds true and by using step 2 you come up with a result that also holds true in step 3 then by induction your identity holds true for $n\geq n_0$ 
