# The sum of the first n odd natural numbers

Is there an easy way to calculate the some of the first n natural numbers?

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

Let's look at the problem for n = 1,2,3,4,5

1=1=1^{2}

1+3=4=2^{2}

1+3+5=9=3^{2}

1+3+5+7=16=4^{2}

1+3+5+7+9=25=5^{2}

So the answer seems to be:

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

*Proof 1:*

We can arrange squares in this way. The big square has n^{2} little squares. We see that the number of little squares is also 1+3+5+...+(2n-1).

*Proof 2:*

We will use induction.

Step 1.

For n = 1 it's true that 1 = 2*1-1

Step 2.

We suppose that 1+3+5+...+(2n-1) = n^{2}

and want to prove that: 1+3+5+...+(2(n+1)-1) = (n+1)^{2}

We add (2(n+1) -1) to this:

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

and get:

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

so:

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

but n^{2}+2n+1 = (n+1)^{2}

so we finally have:

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