We know sum squares of first n natural numbers is .
How to compute sum of squares of first n even natural numbers?
We need to compute 22 + 42 + 62 + …. + (2n)2
EvenSum = 22 + 42 + 62 + .... + (2n)2 = 4 x (12 + 22 + 32 + .... + (n)2) = 4n(n+1)(2n+1)/6 = 2n(n+1)(2n+1)/3
Sum of squares of first 3 even numbers = 2n(n+1)(2n+1)/3 = 2*3(3+1)(2*3+1)/3 = 56 22 + 42 + 62 = 4 + 16 + 36 = 56
How to compute sum of squares of first n odd natural numbers?
We need to compute 12 + 32 + 52 + …. + (2n-1)2
OddSum = (Sum of Squares of all 2n numbers) - (Sum of squares of first n even numbers) = 2n*(2n+1)*(2*2n + 1)/6 - 2n(n+1)(2n+1)/3 = 2n(2n+1)/6 [4n+1 - 2(n+1)] = n(2n+1)/3 * (2n-1) = n(2n+1)(2n-1)/3
Sum of squares of first 3 odd numbers = n(2n+1)(2n-1)/3 = 3(2*3+1)(2*3-1)/3 = 35 12 + 32 + 52 = 1 + 9 + 25 = 35
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