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.

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 ${(k+1)(4(k+1)^2+6(k+1)-1) \over 3}$.
