The sum of first n odd natural numbers is
Question:

The sum of first n odd natural numbers is

(a) 2n − 1

(b) 2n + 1

(c) n2

(d) n2 − 1

Solution:

In this problem, we need to find the sum of first n odd natural numbers.

So, we know that the first odd natural number is 1. Also, all the odd terms will form an A.P. with the common difference of 2.

So here,

First term (a) = 1

Common difference (d) = 2

So, let us take the number of terms as n

Now, as we know,

$S_{n}=\frac{n}{2}[2 a+(n-1) d]$

So, for terms,

$S_{n}=\frac{n}{2}[2(1)+(n-1) 2]$

$=\frac{n}{2}[2+2 n-2]$

$=\frac{n}{2}(2 n)$

$=n^{2}$

Therefore, the sum of first $n$ odd natural numbers is $S_{n}=n^{2}$.

Hence the correct option is (c).