Sum of squares of odd numbers 1^2 + 3^2 + 5^2 + ... + (2n+1)^2

[garyljc]

Homework Statement

Find a formula for 1^2 + 3^2 + 5^2 + ... + (2n+1)^2

The Attempt at a Solution

What i did was Sum of (2n+1)^2
and then using standard formula like 1/6 n(n+1)(2n+1) and 1/2 n (n+1)

but when i insert n=1 it's gives me the 2nd term which is 3^2 = 9
but I'm looking for the summation formula, what I'm looking for is when i insert n=2 , the sum should be 1^2 + 3^2 = 10

but why can't i find it ?
am i heading the right directoin ?

 

[HallsofIvy]

So you want the sum of squares of the odd integers, up to 2n+1?

The sum you are using is the sum of squares of all integers up to n. to get, first, the sum of all integers up to 2n+1, use 1/6 (2n+1)(2n+1+1)(2(2n+1)+1) (the formula you give with 2n+1 in place of n). Once you have that, just subtract off the sum of squares of even numbers up to 2n!

How do you get that? Well, (2)²+ (4)²+ (6)²+ ...(2n)²= 4(1²)+ 4(2²)+ 4(²)+ ...4(n²)= 4(1²+ 2²+ 3²+ ...+ n²) or just 4 times the sum of squares of all integers up to n.