Question

Statement $1$: The variance of first $n$ even natural numbers is $\frac{n^2-1}{4}$
Statement $2$: The sum of first $n$ natural numbers is $\frac{n\left(n+1\right)}{2}$ and the sum of squares of first $n$ natural numbers is $\frac{n(n+1)(2n+1)}{6}$

• A. Statement $1$ is True, Statement $2$ is False
• B. Statement $1$ is False, Statement $2$ is True
• C. Statement $1$ is False, Statement $2$ is False
• D. Statement $1$ is True, Statement $2$ is True

Answer 1

The correct option is B

Statement $1$ is False, Statement $2$ is True

Explanation for correct option :

Step-1: Verify statement $1$

We know that the sum of the first $n$ natural number is $=\frac{n\left(n+1\right)}{2}$

Sum of first $n$ even natural numbers is $=2+4+6+...+2n$

$$\begin{gathered} =2\left(1+2+3+...+n\right) \\ =2\frac{n\left(n+1\right)}{2} \\ =n\left(n+1\right) \end{gathered}$$

The mean$\left(\overline{x}\right)$ of the first $n$ even natural numbers is $=\frac{n\left(n+1\right)}{n}$

$=n+1$

We know that the sum of the square first $n$ natural number is $=\frac{n(n+1)(2n+1)}{6}$

Sum of the square of first $n$ even natural numbers $=2^2+4^2+6^2+...+{\left(2n\right)}^2$

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

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

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

So statement $1$ is incorrect

Step-2: Verify statement $2$

We know that the sum of the first $n$ natural number $=\frac{n\left(n+1\right)}{2}$

We know that the sum of the squares first $n$ natural numbers $=\frac{n(n+1)(2n+1)}{6}$

So statement $2$ is correct

Hence, option(B) i.e. Statement $1$ is False, Statement $2$ is True is correct