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.

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*}
