variance of first n odd natural numbers and

variance of first n even numbers

First n even numbers are,
2, 4, 6, 8.........2n
Now calculating the mean we get, 

$$\begin{gathered} \mu =\frac{2+4+6+8+......+2n}{n} \\ =\frac{2(1+2+3+4+......+n)}{n} \\ =\frac{2\left({\displaystyle \frac{n(n+1)}{2}}\right)}{n}=(n+1) \end{gathered}$$
Now calculating sum of the square of first n even natural numbers is, 
$$\begin{gathered} \sum _{n = 2}^{2n}{n}^2= 2^2+4^2+6^2+........+(2n{)}^2 \\ =2^2\left(1^2+2^2+3^2+........+(n{)}^2\right) \\ =4\left(\frac{n(n+1)(2n+1)}{6}\right) \left[u\sin g the formula{1}^2+2^2+3^2+........+(n{)}^2=\frac{n(n+1)(2n+1)}{6} \right] \\ =\frac{2n(n+1)(2n+1)}{3} \end{gathered}$$
Now calculating variance, 
$$\begin{gathered} =\frac{{\displaystyle \sum _{n = 2}^{2n}}{n}^2}{n}-{\mu }^2 \\ =\frac{2n(n+1)(2n+1)}{3n}-(n+1{)}^2 \\ =\frac{(n+1)(n-1)}{3} \\ =\frac{(n^2-1)}{3} \end{gathered}$$
Similarly you can calculate for the odd numbers , the method will be same only you have to take the values as 1, 3, 5,.......(2n-1).
Hope your doubt is clear now.