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 $$\frac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6}.$$
A.
Statement - 1 is true, Statement - 2 is true Statement - 2 is not a correct explanation for Statement - 1
B.
Statement - 1 is true, Statement - 2 is false.
C.
Statement - 1 is false, Statement - 2 is true.
D.
Statement - 1 is true, Statement - 2 is true.
Statement - 2 is a correct explanation for Statement - 1.
Answer :
Statement - 1 is false, Statement - 2 is true.
Solution :
For the numbers 2, 4, 6, 8, . . . . . ,2$$n$$
$$\eqalign{
& \overline x = \frac{{2\left[ {n\left( {n + 1} \right)} \right]}}{{2n}} = \left( {n + 1} \right) \cr
& {\text{And}}\,\,Var = \frac{{\sum {{{\left( {x - \overline x } \right)}^2}} }}{{2n}} \cr
& = \frac{{\sum {{x^2}} }}{n} - {\left( {\overline x } \right)^2} \cr
& = \frac{{4\sum {{n^2}} }}{n} - {\left( {n + 1} \right)^2} \cr
& = \frac{{4n\left( {n + 1} \right)\left( {2n + 1} \right)}}{{6n}} - {\left( {n + 1} \right)^2} \cr
& = \frac{{2\left( {2n + 1} \right)\left( {n + 1} \right)}}{3} - {\left( {n + 1} \right)^2} \cr
& = \,\left( {n + 1} \right)\,\left[ {\frac{{4n + 2 - 3n - 3}}{3}} \right] \cr
& = \,\frac{{\left( {n + 1} \right)\left( {n - 1} \right)}}{3} \cr
& = \,\frac{{{n^2} - 1}}{3} \cr} $$
∴ Statement - 1 is false. Clearly, statement - 2 is true.