  Question

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

A
Statement 1 is true, Statement2 is true,Statement 2 is a correct explanation for Statement 1  B
Statement 1 is true, Statement2 is true;Statement2 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  Solution

The correct option is D Statement 1 is false, Statement 2 is trueSum of first n even natural numbers$$\displaystyle =2+4+6+...+2n=2(1+2+...+n)$$ $$\displaystyle =2\frac{n(n+1)}{2}=n(n+1)$$ Mean $$\displaystyle (\bar{x})=\frac{n(n+1)}{n}=n+1$$ variance $$\displaystyle =\frac{1}{n}(\sum x_{1})^{2}-(\bar{x})^{2} =\frac{1}{n}(2^{2}+4^{2}+...+(2n)^{2})-(n+1)^{2}$$$$\displaystyle =\frac{1}{n}2^{2}(1^{2}+2^{2}+...+n^{2})-(n+1)^{2}$$$$\displaystyle =\frac{4}{n}\frac{n(n+1)(2n+1)}{6}-(n+1)^{2}$$$$\displaystyle =\frac{2}{3}(n+1)(2n+1)-(n+1)^{2}$$$$\displaystyle =\frac{n+1}{3}[2(2n+1)-3(n+1)]$$$$\displaystyle=\frac{(n+1)}{3} (n-1)=\frac{n^{2}-1}{3}$$Hence statement $$1$$ is false while statement $$2$$ is trueMathematics

Suggest Corrections  0  Similar questions
View More  People also searched for
View More 