Find the sum of first $n$ odd natural numbers.



Given:

First $n$ odd natural numbers.

To do:

We have to find the sum of first $n$ odd natural numbers.

Solution:

First $n$ odd natural numbers are $1, 3, 5, ....., n$.

The above series is in A.P. where $a=1, d=5-3=2$

Sum of $n$ terms of an A.P. $S_n=\frac{n}{2}(2a+(n-1)d)$

Therefore, sum of first $n$ odd natural numbers is,

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

$=\frac{n}{2} \times 2(1+n-1)$

$=n(n)$

$=n^2$

Therefore, the sum of first $n$ odd natural numbers is $n^2$.

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