Question

The weighted mean of first $n$ natural numbers whose weight 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)

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

Weighted mean $=\frac{1.1^2+2.2^2+...+n.n^2}{1^2+2^2+...+n^2}$
$=\frac{\sum n^3}{\sum n^2}=\frac{\frac{n(n+1)}{2}.\frac{n(n+1)}{2}}{\frac{n(n+1)(2n+1)}{6}}=\frac{3n(n+1)}{2(2n+1)}$