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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 =

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

so using formula of sum in AP we get

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

so using S =

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__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