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What is the variance of the first n positive integers?

1. (n² + 1) / 12
2. (n² − 1) / 12
3. (n + 1)(2n + 1) / 6
4. [n(n + 1)]² / 12

Correct answer: (n² − 1) / 12

Solution

For the first n integers, mean=(n+1)/2 and mean of squares=(n+1)(2n+1)/6. Variance = (n+1)(2n+1)/6 - ((n+1)/2)^2 = (n^2-1)/12. The stored option (n+1)(2n+1)/6 is the mean of squares, not the variance, so it is wrong; the correct answer is (n^2-1)/12.

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