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Question: The variance of first \(n\) natural numbers, is A. \(\dfrac{{n + 1}}{2}\) B. \(\dfrac{{{n^2} + ...

The variance of first nn natural numbers, is
A. n+12\dfrac{{n + 1}}{2}
B. n2+112\dfrac{{{n^2} + 1}}{{12}}
C. n2−16\dfrac{{{n^2} - 1}}{6}
D. n2−112\dfrac{{{n^2} - 1}}{{12}}

Explanation

Solution

For required solution, we will see first nn natural numbers and then we have to calculate the variance of those numbers. Now, to calculate variance we need to find mean of all the nn natural numbers. Now, we know, variance can be calculated using the formula σ2=1n∑xi2−(xˉ)2{\sigma ^2} = \dfrac{1}{n}\sum {x_i^2 - {{\left( {\bar x} \right)}^2}} , where xˉ\bar x is the mean of the nn natural numbers and nn is the total count of the numbers.

Complete step-by-step answer:
Now, we know,
The first nn natural numbers are = 1, 2, 3, 4, 5….. nn
The count of these numbers is nn
Now, to find mean we have to use the formula,
Mean, xˉ=∑i=1nxin\bar x = \dfrac{{\sum\limits_{i = 1}^n {{x_i}} }}{n}
So, putting values in the formula,
xˉ=1+2+3+4+5....nn\bar x = \dfrac{{1 + 2 + 3 + 4 + 5....n}}{n}
Now, using formula to find sum of nn natural numbers
s=n(n+1)2s = \dfrac{{n\left( {n + 1} \right)}}{2}
Now, putting the above value in mean,

⇒xˉ=n(n+1)2×n ⇒xˉ=(n+1)2  \Rightarrow \bar x = \dfrac{{n\left( {n + 1} \right)}}{{2 \times n}} \\\ \Rightarrow \bar x = \dfrac{{\left( {n + 1} \right)}}{2} \\\

Now, xˉ2=(n+1)24.............(1){\bar x^2} = \dfrac{{{{\left( {n + 1} \right)}^2}}}{4}.............\left( 1 \right)
Now, we know that the sum of the squares of nn natural numbers is n(n+1)(2n+1)6\dfrac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6}
So, 1n∑xi2=n(n+1)(2n+1)n×6.................(2)\dfrac{1}{n}\sum {x_i^2 = } \dfrac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{{n \times 6}}.................\left( 2 \right)

Now, putting values of equation (1) and (2) in variance formula,
⇒σ2=n(n+1)(2n+1)6n−(n+1)24\Rightarrow {\sigma ^2} = \dfrac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{{6n}} - \dfrac{{{{\left( {n + 1} \right)}^2}}}{4}
Now, on simplifying the equation,

⇒σ2=n(n+1)(2n+1)6n−(n+1)24 ⇒σ2=(n+1)2(2n+13−n+12) ⇒σ2=(n+1)2(4n+2−3n−36) ⇒σ2=(n+1)2(n−16) ⇒σ2=n2−112  \Rightarrow {\sigma ^2} = \dfrac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{{6n}} - \dfrac{{{{\left( {n + 1} \right)}^2}}}{4} \\\ \Rightarrow {\sigma ^2} = \dfrac{{\left( {n + 1} \right)}}{2}\left( {\dfrac{{2n + 1}}{3} - \dfrac{{n + 1}}{2}} \right) \\\ \Rightarrow {\sigma ^2} = \dfrac{{\left( {n + 1} \right)}}{2}\left( {\dfrac{{4n + 2 - 3n - 3}}{6}} \right) \\\ \Rightarrow {\sigma ^2} = \dfrac{{\left( {n + 1} \right)}}{2}\left( {\dfrac{{n - 1}}{6}} \right) \\\ \Rightarrow {\sigma ^2} = \dfrac{{{n^2} - 1}}{{12}} \\\

Hence, the variance of first nn natural numbers is n2−112\dfrac{{{n^2} - 1}}{{12}}

Hence, the correct option for the given problem is D.

Note: In the above question, we have used the formulas of mean and variance. We have also used the properties of arithmetic progression to find the sum of the nn natural numbers. The calculations done in the solution are a bit typical and we cannot make mistakes, so do them precisely.. The calculation of mean is simple but it should also be rechecked twice to avoid any mistakes in these types of questions.