Question

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

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

Answer 1

The correct option is C $\frac{2n+1}{3}$

$Weightedmean=\frac{1.1+2.2+3.3+....+n.n}{1+2+3+.....+n}=\frac{\sum n^2}{\sum n}=\frac{\frac{n(n+1)(2n+1)}{6}}{\frac{n(n+1)}{2}}=\frac{2n+1}{3}$