Calculate the mean deviation about the mean of the set of first $n$ natural numbers when $n$ is an odd number.

The first $n$ natural numbers are $1, 2, 3, \ldots, n$.
The mean $\bar{x} = \frac{1+2+3+\cdots+n}{n} = \frac{n(n+1)}{2n} = \frac{n+1}{2}$.
The mean deviation about the mean is given by $MD = \frac{1}{n} \sum_{i=1}^{n} |x_i - \bar{x}|$.
$MD = \frac{1}{n} \sum_{i=1}^{n} |i - \frac{n+1}{2}|$.
Since $n$ is odd,let $n = 2k+1$. The terms are symmetric around the mean $\frac{n+1}{2}$.
$MD = \frac{2}{n} \left[ \sum_{i=1}^{(n-1)/2} (\frac{n+1}{2} - i) \right] = \frac{2}{n} \left[ \frac{n-1}{2} \cdot \frac{n+1}{2} - \frac{(\frac{n-1}{2})(\frac{n+1}{2})}{2} \right]$.
$MD = \frac{2}{n} \left[ \frac{n^2-1}{8} + \frac{n^2-1}{8} \right] = \frac{2}{n} \left[ \frac{n^2-1}{4} \right] = \frac{n^2-1}{4n}$.