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Question: The weighted mean of first n natural numbers whose weights are equal to the squares of corresponding...

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

A

n+12\frac{n + 1}{2}

B

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

C

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

D

n(n+1)2\frac{n(n + 1)}{2}

Answer

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

Explanation

Solution

Weighted mean = 1.12+2.22+......+n.n212+22+......+n2\frac{1.1^{2} + 2.2^{2} + ...... + n.n^{2}}{1^{2} + 2^{2} + ...... + n^{2}}

=Σn3Σn2=n(n+1)2n(n+1)2n(n+1)(2n+1)6=3n(n+1)2(2n+1)= \frac{\Sigma n^{3}}{\Sigma n^{2}} = \frac{\frac{n(n + 1)}{2}\frac{n(n + 1)}{2}}{\frac{n(n + 1)(2n + 1)}{6}} = \frac{3n(n + 1)}{2(2n + 1)}.