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Compute \(1^2 + 3^2 + 5^2 + \dots + 99^2\)

 Jul 12, 2020
 #1
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Use "Sum of consecutive squares" formula to sum them up:

 

n =99; a = 1/6 * (2*n + 1)*n*(n + 1) =328,350

 Jul 12, 2020
 #2
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The numbers aren't consecutive, they increase by 2, so:

 

 Jul 13, 2020

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