What is the difference between the sum of first n odd natural numbers and that of first n natural numbers?
$\dfrac{n}{2}$ +1
$\dfrac{n(n-1)}{2}$
$\dfrac{n(n+1)}{2}$
$\dfrac{n(n+1)(2n+1}{6}$
Explanation:
$sum of first `n' odd nos=n^{2}$
$sum of first `n' natural nos=\dfrac{n(n+1)}{2}$
$Difference =n^{2}-[\dfrac{n^{2}+n}{2}]^{2}$
$=\dfrac{2 n^{2}-n^{2}-n}{2}$
$=\dfrac{n^{2}-n}{2}$
$=\dfrac{n(n-1)}{2}$