JEE Main · 2009 · 12 Apr 2009 · Offline · Statistics

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 ( n + 1 )}{2}

and the sum of squares of first

n

natural numbers is

\frac{n ( n + 1 ) ( 2 n + 1 )}{6}

(A)Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1
(B)Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1
(C)Statement-1 is true, Statement-2 is false
(D)Statement-1 is false, Statement-2 is true✓ Correct

Answer: (D)

Solution

The first

n

even natural numbers are

2, 4, \dots, 2 n

, summing to

n (n + 1)

so the mean is

n + 1

. Their sum of squares is

4 (1^{2} + \dots + n^{2}) = \frac{2 n ( n + 1 ) ( 2 n + 1 )}{3}

, and the variance is

\frac{1}{n}

of this minus the square of the mean,

\frac{2 ( n + 1 ) ( 2 n + 1 )}{3} - (n + 1)^{2} = \frac{( n + 1 ) ( n - 1 )}{3} = \frac{n ^{2} - 1}{3}

. This is

\frac{n ^{2} - 1}{3}

, not

\frac{n ^{2} - 1}{4}

, so the first statement is false while the standard summation formulas of the second statement are true.

JEE Main · 2009 · 12 Apr 2009 · Offline · Single correct option