logo

Ask Questions, Get Answers

X
 
Home  >>  JEEMAIN and NEET  >>  Mathematics  >>  Class11  >>  Statistics

The A.M of squares of first $n$ natural numbers is

$\begin {array} {1 1} (A)\;\large\frac{n+1}{6} & \quad (B)\;\large\frac{n^2-1}{6} \\ (C)\;\large\frac{(n+1)(2n+1)}{6} & \quad (D)\;None\: of \: these \end {array}$

Download clay6 mobile app

1 Answer

$ \overline x = \large\frac{1}{n} $ $ \Sigma x = \large\frac{1^2+2^2+.....n^2}{n}$
$ = \large\frac{n(n+1)(2n+1)}{6.n}$
$ = \large\frac{(n+1)(2n+1)}{6}$
Ans : (C)
answered Jan 25, 2014 by thanvigandhi_1
 

Related questions

Ask Question
Tag:MathPhyChemBioOther
...
X