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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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