If the mean of n observations 1^2, 2^2, 3^2, | vedclass

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(A) The mean of the first $n$ squares is given by $\frac{\sum_{i=1}^{n} i^2}{n} = \frac{n(n+1)(2n+1)}{6n} = \frac{(n+1)(2n+1)}{6}$.Given that the mean

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$11$

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(A) The mean of the first $n$ squares is given by $\frac{\sum_{i=1}^{n} i^2}{n} = \frac{n(n+1)(2n+1)}{6n} = \frac{(n+1)(2n+1)}{6}$.Given that the mean is $\frac{46n}{11}$,we have:$\frac{(n+1)(2n+1)}{6} = \frac{46n}{11}$$11(2n^2 + 3n + 1) = 276n$$22n^2 + 33n + 11 = 276n$$22n^2 - 243n + 11 = 0$Factoring the quadratic equation:$(n - 11)(22n - 1) = 0$Since $n$ must be a positive integer,$n = 11$.

If the mean of $n$ observations $1^2, 2^2, 3^2, \dots, n^2$ is $\frac{46n}{11}$,then what is the value of $n$?

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