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

Example:

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

Example:

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

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above

Attention reader! Don’t stop learning now. Get hold of all the important CS Theory concepts for SDE interviews with the CS Theory Course at a student-friendly price and become industry ready.