Question

The reciprocal of the weighted mean of first $n$ natural numbers whose weights are equal to the squares of the corresponding number is

• A. $\frac{2(2n+1)}{3n(n+1)}$
• B. $\frac{3n(n+1)}{n(2n+1)}$
• C. $\frac{3n(n+1)}{(2n+1)}$
• D. none of these

Answer 1

The correct option is A $\frac{2(2n+1)}{3n(n+1)}$

$x$	$1$	$2$	$3$	$4$	$...n$
$f_w$	$1$	$2^2$	$3^2$	$4^2$	$...n^2$

Weighted mean $=\frac{w_1{x}_1+w_2{x}_2+...+w_i{x}_i}{w_1+w_2+...+w_i}=\frac{\sum w_i{x}_i}{\sum w_i}$

$w\left(\overline{x}\right)=$ weighted mean $=\frac{1.1+2.2^2+3.3^2+...+n.n^2}{1^2+2^2+3^2+...+n^3}$

$=\frac{1^3+2^3+3^3+...+n^3}{1^2+2^2+3^2+...+n^2}$$=\frac{n^2{(n+1)}^2}{4}\times \frac{6}{n(n+1)(2n+1)}=\frac{3n(n+1)}{2(2n+1)}$

$\therefore$ The reciprocal of the weighted mean is $\frac{2(2n+1)}{3n(n+1)}$