Question

Consider the following two statements :

Statement $1$ : The variance of first $n$ even natural numbers is $\frac{n^2-1}{4}$.

Statement $\text{2}$ : The sum of first $n$ natural number is $\frac{n\left(n+1\right)}{2}$ and the sum of the squares of first $n$ natural numbers is $\frac{n\left(n+1\right)\left(2n+1\right)}{6}$.

Then which of one of the following choices is correct?

• A. Both statement $1$ and statement $\text{2}$ are correct and statement $\text{2}$ is a correct explanation for statement $1$
• B. Both statement $1$ and statement $\text{2}$ are correct and statement $\text{2}$ is an incorrect explanation for statement $1$
• C. Statement$1$ is correct and statement $\text{2}$ is incorrect.
• D. Statement$1$ is incorrect, statement $\text{2}$ is correct

Answer 1

The correct option is D

Statement$1$ is incorrect, statement $\text{2}$ is correct

Explanation for the correct option :

Finding the true statement:

We know that the sum of the first $n$natural numbers is $\frac{n\left(n+1\right)}{2}$ and the sum of the squares of first $n$ natural numbers is $\frac{n\left(n+1\right)\left(2n+1\right)}{6}$.

So, Statement-2 is correct.

The sum of the first $n$even natural numbers

$$\begin{gathered} =2+4+\dots +2n \\ =2\left(1+2+\dots +n\right) \\ =2\times \frac{n\left(n+1\right)}{2}\left[\because 1+2+\dots +n=\frac{n\left(n+1\right)}{2}\right] \\ =n\left(n+1\right) \end{gathered}$$

So, the mean of the first $n$ even natural numbers

$$\begin{gathered} \text{Mean}\left(\overline{x}\right)=\left(\frac{n\left(n+1\right)}{n}\right)\left(\text{Mean}=\frac{\text{Sum}}{\text{Number}}\right) \\ =n+1 \end{gathered}$$

Here, we will use the formula for variance i.e. $\text{Variance}=\frac{1}{n}\sum _{i=1}^n{\left(x_i\right)}^2-{\left(\overline{x}\right)}^2$, where $x_1,x_2,\dots ,x_n$ are the given numbers and $\overline{x}$ is the mean of the given numbers i.e. $\overline{x}=\frac{x_1+x_2+\dots +x_n}{n}$.

Hence, the variance of the first $n$even natural numbers is

$$\begin{gathered} \text{Variance}=\frac{1}{n}\left(2^2+4^2+\dots +4n^2\right)-{\left(\overline{x}\right)}^2 \\ =\frac{1}{n}\times 2^2\left(1^2+2^2+\dots +n^2\right)-{(n+1)}^2 \\ =\frac{4}{n}\times \frac{n(n+1)(2n+1)}{6}-{(n+1)}^2(\text{sum of squares of}n\text{natural numbers}) \\ =\frac{(n+1)(4n+2-3n-3)}{3} \\ =\frac{(n+1)(n-1)}{3} \\ =\frac{n^2-1}{3} \end{gathered}$$

Hence, Statement 1: The variance of first $n$ even natural numbers is $\frac{n^2-1}{4}$ is incorrect.

Therefore, option (D) is the correct answer.