$\begin{array}{1 1}(A)\;\sqrt {\large\frac{n^2-1}{12}}\\(B)\;\large\frac{n^2-1}{4n}\\(C)\;0\\(D)\;\large\frac{n-1}{12}\end{array} $

mean of first n natural numbers where n is odd number $\large\frac{n+1}{2}$

mean deviation $= \large\frac{\sum |x_i -x_n|}{n}$

$\qquad= \large\frac{1}{n} \bigg[ \bigg| 1- \frac{n}{2} \bigg| $$+ \bigg|\bigg(2- \frac{n}{2} +1 \bigg)\bigg|.........\bigg|\bigg(n- \large\frac{n+1}{2} \bigg)\bigg|\bigg]$

$\qquad= \large\frac{2}{n}$$ \large\frac{\bigg [ \bigg (\large\frac{ n-1}{2}\bigg) \bigg( \large \frac{n-1}{2} +1 \bigg)\bigg]} {2}$

$\qquad= \large\frac{ (n-1)(n+1)}{4n}$

$\qquad= \large\frac{n^2 -1}{4n}$

Hence B is the correct answer.

please explain the 2nd and 3rd step

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