Prove the identity 1+3+5+...+(2n-1)=n^2

For demonstrating the identity, we'll use the method of
mathematical induction, which consists in 3
steps:

1) verify that the method works for the number
1;

2) assume that the method works for an arbitrary number,
k;

3) prove that if the method works for an arbitrary
number k, then it work for the number k+1, too.

4) after
the 3 steps were completed, then the formula works for any
number.

Now, we'll start the first
step:

1) 1=1^2 => 1=1
true.

2) 1+3+5+...+(2k-1)=k^2 ,
true.

3) If 1+3+5+...+(2k-1)=k^2,
then

1+3+5+...+(2k-1)+(2k+2-1)=(k+1)^2

Let's
see if it is true.

For the beginning, we notice that the
sum from the left contains the assumed true equality, 1+3+5+...+(2k-1)=k^2. So, we'll
re-write the sum by substituting 1+3+5+...+(2k-1) with
k^2.

k^2 + (2k+1) =
(k+1)^2

We'll open the
brackets:

k^2 + 2k+1 = k^2 + 2k+1
true.

4) The 3 steps were completed, so the identity is tru
for any value of
n.

1+3+5+...+(2n-1)=n^2