The nth odd positive integer is given by either

2n+1, where n begins with 0

or by

2n-1, where n begins with 1

I will use the latter, since ordinarily we begin counting with 1, not 0. 

Let Sn represent the sum.

Sn = 1 + 3 + 5 + ··· + (2n-1)

The odd number just before (2n-1) is (2n-3) and the one before that is
(2n-5), etc. So the above series can be written

Sn = 1 + 3 + 5 + ··· + (2n-5) + (2n-3) + (2n-1)

This is the same sum as when we reverse the terms:

Sn = (2n-1) + (2n-3) + (2n-5) + ··· + 5 + 3 + 1

Now when we add the two equations term by term, we get:

2Sn = [1+(2n-1)]+[3+(2n-3)]+[5+(2n-5)]+ ··· +[(2n-5)+5]+[(2n-3)+3]+[(2n-1)+1]

or upon simplifying

2Sn = 2n + 2n + 2n + ··· + 2n + 2n + 2n

Factoring 2 out of the right side:

2Sn = 2(n + n + n + ··· + n + n + n)

Dividing both sides by 2

Sn = n + n + n + ··· + n + n + n

There are n terms on the right, all which are n, so their sum is n×n or n².

Therefore

Sn = n²

And we see that

1 = 1 = 1²

1+3 = 4 = 2²

1+3+5 = 9 = 3²

1+3+5+7 = 16 = 4²

etc.

Edwin