The sum of the first n odd natural numbers is equal to n2

Step 1 
Let 1, 3, 5, 7, 9, n odd numbers be the first.
It is an Arithmatic Progression, whose first term ${\displaystyle \left(a\right)=1}$, and common difference (d) is 2. 
Step 2 
The total of these n terms ${\displaystyle A\cdot P=\frac{n}{2}[2a+(n-1)d]}$ 
first n odd numbers added together ${\displaystyle =\frac{n}{2}[2\times 1+(n-1)\times 2]}$ 
${\displaystyle =\frac{n}{2}[2+2n-2]}$ 
${\displaystyle =\frac{n}{2}\times n=n^2}$ 
Hence, the sum of first n odd natural numbers is equal to ${\displaystyle n^2}$. Hence, proved.

Proving by example
Sum of first 3 odd numbers is $3^2=9$
First 3 odd numbers are 1+3+5=9
Hence ,Sum of first n odd natural numbers is $n^2$