Question

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

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

Answer 1

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

The corresponding weights of $1,2,3,\mathrm{4...}n$ are $1,2,3,\mathrm{4...}n$

hence weighted average = $\frac{1\times 1+2\times 2+3\times 3+4\times 4+....}{1+2+3+\mathrm{4...}+n}$

= $\frac{1^2+2^2+3^2+....+n^2}{1+2+3+...+n}$

= $\frac{\frac{n(n+1)(2n+1)}{6}}{\frac{n(n+1)}{2}}$

= $\frac{2n+1}{3}$