| 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 \[\frac{n(n+1)(2n+1)}{6}.\] |
A) Statement-1 is true, statement-2 is true, statement-2 is 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 :
Variance of first n even natural numbers. \[=\left( \frac{{{2}^{2}}+{{4}^{2}}+....+{{(2n)}^{2}}}{n} \right)-{{\left( \frac{2+4+.....+2n}{n} \right)}^{2}}\]\[=\left( \frac{4(n)\,(n+1)\,(2n+1)}{6n}-\,\frac{4{{n}^{2}}{{(n+1)}^{2}}}{4{{n}^{2}}} \right)\] \[\frac{2}{3}\,\left( 2{{n}^{2}}+3n+1 \right)-\left( {{n}^{2}}+2n+1 \right)\,=\frac{{{n}^{2}}-1}{3}\]
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