Question

The weighted mean of first $n$ natural numbers, whose weights are proportional to the corresponding numbers is $\frac{2n+1}{3}$.Is it true? If true enter 1 else 0.

Answer 1

$Letweightofnbe\lambda n,where\lambda =n,n\in Nthenweightedmean$

$=\left(\frac{(1\times \lambda )+(2\times \lambda )+(3\times \lambda )+.......+(n\times \lambda )}{1\lambda +2\lambda +3\lambda +......+n\lambda }\right)$

$=\left(\frac{1^2+2^2+3^2+.....+n^2}{1+2+3+....+n}\right)$

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

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