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Question

Mathematics Question on Statistics

The weighted mean of first n natural numbers, whose weights are proportional to the corresponding numbers, is

A

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

B

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

C

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

D

2n+13\frac{2n +1}{3}

Answer

2n+13\frac{2n +1}{3}

Explanation

Solution

Here, Required mean =λ(12+22+32+...+n2)λ(1+2+3+...+n)= \frac{\lambda\left(1^{2} + 2^{2}+3^{2} +...+n^{2}\right)}{\lambda\left(1+2+3+...+n\right)} =2λn(n+1)(2n+1)6λn(n+1)= \frac{2\lambda n \left(n+1\right)\left(2n+1\right)}{6\lambda n \left(n+1\right)}