Question

The variance of first $n$ natural numbers

• A. $\frac{n\left(n+1\right)}{2}$
• B. $\frac{\left(n+1\right)\left(n+5\right)}{12}$
• C. $\frac{\left(n+1\right)\left(n-5\right)}{12}$
• D. $\frac{n^2-1}{12}$

Answer 1

The correct option is D

$\frac{n^2-1}{12}$

Explanation for correct option:

Step 1: Calculate the individual term

The first $n$ natural number is $1,2,3,...,n$

$$\begin{gathered} \therefore \sum _{i=1}^{n}i=\frac{n\left(n+1\right)}{2} \\ \sum _{i=1}^n{i}^2=\frac{n\left(n+1\right)\left(2n+1\right)}{6} \end{gathered}$$

Step 2: Calculate the Variance

Variance$=\frac{{\displaystyle \sum _{i=1}^n}{i}^2}{n}-{\left(\frac{{\displaystyle \sum _{i=1}^n}i}{n}\right)}^2$

$$\begin{gathered} =\frac{n\left(n+1\right)\left(2n+1\right)}{6n}-{\left(\frac{n\left(n+1\right)}{2n}\right)}^2 \\ =\frac{\left(n+1\right)\left(2n+1\right)}{6}-\frac{{\left(n+1\right)}^2}{4} \\ =\frac{\left(n+1\right)\left[2\left(2n+1\right)-3\left(n+1\right)\right]}{12} \\ =\frac{\left(n+1\right)\left(n-1\right)}{12} \\ =\frac{n^2-1}{12} \end{gathered}$$

Hence, option D is correct.