Find the weighted mean of the first n natural | vedclass

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(B) The weighted mean is given by $\bar{x}_w = \frac{\sum_{i=1}^{n} i \cdot i^2}{\sum_{i=1}^{n} i^2}$.This simplifies to $\bar{x}_w = \frac{\sum_{i=1}

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

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(B) The weighted mean is given by $\bar{x}_w = \frac{\sum_{i=1}^{n} i \cdot i^2}{\sum_{i=1}^{n} i^2}$.This simplifies to $\bar{x}_w = \frac{\sum_{i=1}^{n} i^3}{\sum_{i=1}^{n} i^2}$.Using the standard summation formulas $\sum i^3 = [\frac{n(n+1)}{2}]^2$ and $\sum i^2 = \frac{n(n+1)(2n+1)}{6}$,we get:$\bar{x}_w = \frac{[n(n+1)/2]^2}{n(n+1)(2n+1)/6} = \frac{n^2(n+1)^2}{4} \cdot \frac{6}{n(n+1)(2n+1)}$.Simplifying the expression,we obtain $\bar{x}_w = \frac{3n(n+1)}{2(2n+1)}$.

Find the weighted mean of the first $n$ natural numbers,where the weights are equal to their squares.

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