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

Options

A. `(n+1)/(2)`
B. `(3n(n+1))/(2(2n+1))`
C. `((n+1)(2n+1))/(6)`
D. `(n(n+1))/(2)`

Answer

B

Explanation

Weighted mean `=(1.1^(2)+2.2^(2)+..+n.n^(2))/(1^(2)+2^(2)+..+n^(2))`
`=(sumn^(3))/(sumn^(2))=((n(n+1))/(2)(n(n+1))/(2))/((n(n+1)(2n+1))/(6))=(3n(n+1))/(2(2n+1))`