Question

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

Answer 1

weighted mean $=\frac{{1.1}^2+{2.2}^2+...n.n^2}{1^2+2^2+...+n^2}=\frac{\sum n^3}{\sum n^2}=\frac{\frac{n^2(n+1{)}^2}{4}}{\frac{n(n+1)(2n+1)}{6}}$

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