Question

Statement – 1: The variance of first n even natural number is $\dfrac{{{n^2} - 1}}{4}$
Statement – 2: The sum of first n natural number is $\dfrac{{n\left( {n + 1} \right)}}{2}$ and the sum of squares of first n natural number is $\dfrac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6}$.
$\left( a \right)$ Statement – 1 is true, statement – 2 is true; statement – 2 is a correct explanation for statement – 1.
$\left( b \right)$ Statement – 1 is true, statement – 2 is true; statement – 2 is not a correct explanation for statement – 1.
$\left( c \right)$ Statement – 1 is true, statement – 2 is false.
$\left( d \right)$ Statement – 1 is false, statement – 2 is true.

Answer 1

Hint: In this particular question use the concept that the sum of an A.P series is given as ${S_n} = \dfrac{n}{2}\left( {2a + \left( {n - 1} \right)d} \right)$, where symbols have their usual meanings and the sum of squares of first n natural number is $\dfrac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6}$, so use this concept to reach the solution of the question.

Complete step-by-step solution:
The first natural numbers are given below,
$1, 2, 3, 4, ...........n$
The sum of the first natural numbers are
$1 + 2 + 3 + 4 + ............... + n$
As the above series formed an A.P series, with first term (a) = 1, common difference $(d) = (2 - 1) = (3 – 2) = 1$, and the number of terms is n.
So according to the formula of the sum of an A.P series which is given as,
$ \Rightarrow {S_n} = \dfrac{n}{2}\left( {2a + \left( {n - 1} \right)d} \right)$
Now substitute the values we have,
$ \Rightarrow {S_n} = \dfrac{n}{2}\left( {2\left( 1 \right) + \left( {n - 1} \right)\left( 1 \right)} \right) = \dfrac{n}{2}\left( {2 + n - 1} \right) = \dfrac{{n\left( {n + 1} \right)}}{2}$
So the sum of first n natural numbers is $\dfrac{{n\left( {n + 1} \right)}}{2}$.......................... (1)
And we all know that the sum of squares of the first n natural number is $\dfrac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6}$.
\[ \Rightarrow {1^2} + {2^2} + {3^2} + ........ + {n^2} = \dfrac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6}\]................ (2)
Hence statement – 2 is true.
Statement – 1: The variance of first n even natural number is $\dfrac{{{n^2} - 1}}{4}$
As we know that the even number is always divided by 2.
So the set of even natural numbers are $(2, 4, 6, 8, ........, 2n)$
so the sum of first n even natural numbers are
$S = 2 + 4 + 6 + 8 + ....... + 2n$
Therefore, $S = 2 (1 + 2 + 3 + ......+ n)$
Therefore, $S = 2$ (sum of first n natural numbers)
$ \Rightarrow S = 2\left( {\dfrac{{n\left( {n + 1} \right)}}{2}} \right) = n\left( {n + 1} \right)$
Now as we know that the mean ($\bar x$) is the ratio of sum of numbers to the total numbers.
So the mean of first n even natural numbers is, $\bar x = \dfrac{{n\left( {n + 1} \right)}}{n} = n + 1$....................... (3)
Now as we know that the variance (${\sigma ^2}$) is given as,
$ \Rightarrow {\sigma ^2} = \dfrac{1}{n}{\left( {\sum\limits_{i = 1}^n {{X_i}} } \right)^2} - {\left( {\bar x} \right)^2}$
$ \Rightarrow {\sigma ^2} = \dfrac{1}{n}\left( {{2^2} + {4^2} + {6^2} + ......{{\left( {2n} \right)}^2}} \right) - {\left( {\bar x} \right)^2}$
Now the above expression is also written as,
$ \Rightarrow {\sigma ^2} = \dfrac{1}{n}{2^2}\left( {{1^2} + {2^2} + {3^2} + ......{{\left( n \right)}^2}} \right) - {\left( {\bar x} \right)^2}$
Now substitute the values from equation (2) and (3) in the above equation we have,
$ \Rightarrow {\sigma ^2} = \dfrac{1}{n}{2^2}\left( {\dfrac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6}} \right) - {\left( {n + 1} \right)^2}$
Now simplify it we have,
$ \Rightarrow {\sigma ^2} = \dfrac{2}{3}\left( {\left( {n + 1} \right)\left( {2n + 1} \right)} \right) - {\left( {n + 1} \right)^2}$
$ \Rightarrow {\sigma ^2} = \left( {n + 1} \right)\left[ {\dfrac{2}{3}\left( {2n + 1} \right) - n - 1} \right]$
$ \Rightarrow {\sigma ^2} = \left( {n + 1} \right)\left[ {\dfrac{{4n}}{3} - n + \dfrac{2}{3} - 1} \right]$
$ \Rightarrow {\sigma ^2} = \left( {n + 1} \right)\left[ {\dfrac{n}{3} - \dfrac{1}{3}} \right]$
$ \Rightarrow {\sigma ^2} = \dfrac{{\left( {n + 1} \right)\left( {n - 1} \right)}}{3} = \dfrac{{{n^2} - 1}}{3}$
So this is the required variance of first n even natural numbers.
Hence statement – 1 is false.
So this is the required answer.
Hence option (d) is the correct answer.

Note: Whenever we face such types of question the key concept we have to remember is the formula of the variance which is given as, ${\sigma ^2} = \dfrac{1}{n}{\left( {\sum\limits_{i = 1}^n {{X_i}} } \right)^2} - {\left( {\bar x} \right)^2}$ , where n is the number of terms in the series, $\left( {\sum\limits_{i = 1}^n {{X_i}} } \right)$ is the sum of the series and $\left( {\bar x} \right)$ is the mean of the series.