Question

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

• A. $\frac{n+1}{2}$
• B. $\frac{3n(n+1)}{2(2n+1)}$
• C. $\frac{(n+1)(2n+1)}{6}$
• D. $\frac{n(n+1)}{2}$

Answer 1

Answer: (B)

$\frac{3n(n+1)}{2(2n+1)}$

Here $x_i=1,2,3,4,..................n$
and $w_i=1^2,2^2,3^2,4^2,..........n^2$
$\therefore$ weighted mean $=\frac{\sum w_i{x}_i}{\sum w_i}=\frac{\sum n^3}{\sum n^2}=\frac{\frac{1}{4}\left\{n\right(n+1){\}}^2}{\frac{1}{6}n(n+1)(2n+1)}=\frac{3n(n+1)}{2(2n+1)}$