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=12
1+3=4=22
1+3+5=9=32
1+3+5+7=16=42
1+3+5+7+9=25=52

So the answer seems to be:
1+3+5+...+(2n-1) = n2

Proof 1:

We can arrange squares in this way. The big square has n2 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) = n2
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) = n2

and get:
1+3+5+...+(2n-1) + (2(n+1) -1) = n2 + (2(n+1) -1)

so:
1+3+5+...+(2(n+1) -1) = n2 + 2n+2 -1

but n2+2n+1 = (n+1)2

so we finally have:
1+3+5+...+(2(n+1)-1) = (n+1)2

Average: 4.1 (444 votes)

Back to top