# Mathematics | Sum of squares of even and odd natural numbers?

We know sum squares of first n natural numbers is n(n+1)(2n+1)/6.

**How to compute sum of squares of first n even natural numbers?**

We need to compute 2^{2} + 4^{2} + 6^{2} + …. + (2n)^{2}

EvenSum = 2^{2}+ 4^{2}+ 6^{2}+ .... + (2n)^{2}= 4 x (1^{2}+ 2^{2}+ 3^{2}+ .... + (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 1^{2} + 3^{2} + 5^{2} + …. + (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 1^{2}+ 3^{2}+ 5^{2}= 1 + 9 + 25 = 35

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

Disclaimer: This content belongs to geeksforgeeks, source: http://geeksforgeeks.org