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The sum of the squares of the first \(n\) positive integers \(1^2+2^2+3^2+\ldots+n^2\) is \(n\dfrac{(n+1)(2n+1)}{6}.\)

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The sum of the squares of the first \(n\) positive integers \(1^2+2^2+3^2+\ldots+n^2\) is \(n\dfrac{(n+1)(2n+1)}{6}.\)

by AAPL » Mon May 18, 2026 5:45 pm

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The sum of the squares of the first \(n\) positive integers \(1^2+2^2+3^2+\ldots+n^2\) is \(n\dfrac{(n+1)(2n+1)}{6}.\) What is the sum of the squares of the first \(9\) positive integers?

A. \(90\)
B. \(125\)
C. \(200\)
D. \(285\)
E. \(682\)

OA D
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