Question

Variance of first $n$ natural numbers.
A. $\dfrac{{{n^2} - 1}}{6}$
B. $\dfrac{{{n^2} + 1}}{6}$
C. $\dfrac{{{n^2} - 1}}{{12}}$
D. $\dfrac{{{n^2} + 1}}{{12}}$

Answer 1

Hint: Here in this problem we have to find the variance of the first $n$ natural numbers. Here the natural numbers are the numbers which start from 1 and the numbers here are only integers starting from 1. And also we have to know what variance is, the variance is defined as the average of the squared differences from the mean. Variance is a statistical measurement, it measures how far each number varies in the set of observations from the mean.

Complete step-by-step solution:
Suppose given a set of observations ${x_1},{x_2},{x_3},.....{x_n}$, then the variance denoted by ${\sigma ^2}$is given by:
$ \Rightarrow {\sigma ^2} = \dfrac{{\sum\limits_{i = 1}^n {{x_i}^2} }}{n} - {\left( {\dfrac{{\sum\limits_{i = 1}^n {{x_i}} }}{n}} \right)^2}$
Now the first $n$natural numbers are $1,2,3,......n$
The variance of these first $n$natural numbers is given by:
$ \Rightarrow {\sigma ^2} = \dfrac{{\sum {{n^2}} }}{n} - {\left( {\dfrac{{\sum n }}{n}} \right)^2}$
Here the sum of the first $n$natural numbers is given by:
\[ \Rightarrow \sum n = \dfrac{{n(n + 1)}}{2}\]
The sum of the squared first $n$natural numbers is given by:
\[ \Rightarrow \sum {{n^2}} = \dfrac{{n(n + 1)(2n + 1)}}{6}\]
Now substituting these expressions in the variance expression, which is given below:
$ \Rightarrow {\sigma ^2} = \dfrac{{\sum {{n^2}} }}{n} - {\left( {\dfrac{{\sum n }}{n}} \right)^2}$
$ \Rightarrow {\sigma ^2} = \dfrac{{n(n + 1)(2n + 1)}}{{6n}} - {\left( {\dfrac{{n(n + 1)}}{{2n}}} \right)^2}$
$ \Rightarrow {\sigma ^2} = \dfrac{{(n + 1)(2n + 1)}}{6} - \dfrac{{{{(n + 1)}^2}}}{4}$
$ \Rightarrow {\sigma ^2} = (n + 1)\left( {\dfrac{{2n + 1}}{6} - \dfrac{{n + 1}}{4}} \right)$
On further simplification:
$ \Rightarrow {\sigma ^2} = (n + 1)\left( {\dfrac{{4n + 2 - 3n - 3}}{{12}}} \right)$
$ \Rightarrow {\sigma ^2} = (n + 1)\left( {\dfrac{{n - 1}}{{12}}} \right)$
$ \Rightarrow {\sigma ^2} = \dfrac{{({n^2} - 1)}}{{12}}$
$\therefore {\sigma ^2} = \dfrac{{({n^2} - 1)}}{{12}}$
The variance of the first $n$ natural numbers is $\dfrac{{({n^2} - 1)}}{{12}}$

Option C is the correct answer.

Note: Note that the variance is also given by the ratio of the sum of the squares of deviation from the mean to the total no. of observations minus one, which is the same as the variance formula used here which the difference of the ratio of sum of the squares of all observations to the total no. of observations and the whole square of the mean expression.