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

$\begin {array} {1 1} (A)\;\large\frac{n+1}{2} & \quad (B)\;\large\frac{3n(n+1)}{2(2n+1)} \\ (C)\;\large\frac{(n+1)(2n+1)}{6} & \quad (D)\;\large\frac{n(n+1)}{2} \end {array}$

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Weighted mean = $ \large\frac{1.1^2+2.2^2+....+n.n^2}{1^2+2^2+....+n^2}$
$= \large\frac{\Sigma n^3}{\Sigma n^2} = \large\frac{\large\frac{n^2(n+1)^2}{4}}{\large\frac{n(n+1)(2n+1)}{6}}$
$ = \large\frac{3n(n+1)}{2(2n+1)}$
Ans : B
answered Jan 25, 2014 by thanvigandhi_1

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