The arithmetic mean of the sum of squares of the first n natural numbers is (n+1)(2n+1)/6.
Here's a breakdown of how we arrive at this result:
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Sum of Squares of First n Natural Numbers: The sum of the squares of the first n natural numbers is given by the formula:
$$\sum_{i=1}^{n} i^2 = \frac{n(n+1)(2n+1)}{6}$$
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Arithmetic Mean: To find the arithmetic mean, we divide the sum by the number of terms, which in this case is 'n'.
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Calculation: Therefore, the arithmetic mean is:
$$\text{Arithmetic Mean} = \frac{\frac{n(n+1)(2n+1)}{6}}{n} = \frac{n(n+1)(2n+1)}{6n}$$
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Simplification: Simplifying the expression by cancelling out 'n', we get:
$$\text{Arithmetic Mean} = \frac{(n+1)(2n+1)}{6}$$
Therefore, the arithmetic mean of the sum of squares of the first n natural numbers is (n+1)(2n+1)/6.
