P(n) =1^2+3^2+5^2+...+(2n-1)^2 = n(2n-1)(2n+1)/3

Saluranedukasion 2023-12-03T04:43:41.000Z

P(n) =1^2+3^2+5^2+...+(2n-1)^2 = n(2n-1)(2n+1)/3
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P(n) =1^2+3^2+5^2+...+(2n-1)^2 = n(2n-1)(2n+1)/3