The average of the squares of the first ten n | vedclass

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Question

(D) The sum of the squares of the first $n$ natural numbers is given by the formula: $\frac{n(n+1)(2n+1)}{6}$.The average of the squares of the first

Vedclass · Accepted Answer

$38.5$

Vedclass · Answer

(D) The sum of the squares of the first $n$ natural numbers is given by the formula: $\frac{n(n+1)(2n+1)}{6}$.The average of the squares of the first $n$ natural numbers is the sum divided by $n$:$	ext{Average} = \frac{n(n+1)(2n+1)}{6n} = \frac{(n+1)(2n+1)}{6}$.For the first $10$ natural numbers,substitute $n = 10$ into the formula:$	ext{Average} = \frac{(10+1)(2 	imes 10 + 1)}{6} = \frac{11 	imes 21}{6}$.$	ext{Average} = \frac{231}{6} = 38.5$.

The average of the squares of the first ten natural numbers is

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