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