Question

find the sum of odd natural numbers?

Answer 1

so odd natural numbers are 1 , 3 ,5 ,7 ,9 ............. n terms

so using formula of sum in AP we get

S = 1 + 3 + 5 + ..... n terms

so using S = $\frac{n}{2}[2a+(n-1)d]$

so where a = first term 

n = no of terms 

d = common difference

here the common difference is 3 -1 = 2 = d

first term is 1 

so s = n/2{2 + (n-1)2}

⇒ s = n(1 + n - 1 )     cancelling 2

⇒ s = n² ANSWER

Answer 2

The sum of the first n odd natural numbers 1+3+5+...+(2(n+1)-1) = (n+1)2  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+...+(2n-1) + (2(n+1) -1) = n2 + (2(n+1) -1) so: 1+3+5+...+(2n-1) = n2 and get: and want to prove that: 1+3+5+...+(2(n+1)-1) = (n+1)2 We add (2(n+1) -1) to this:
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