# Sequence formula help

1. Oct 7, 2008

### garyljc

1. The problem statement, all variables and given/known data
Find a formula for 1^2 + 3^2 + 5^2 + .... + (2n+1)^2

3. 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 ?

2. Oct 7, 2008

### HallsofIvy

Staff Emeritus
Re: Sequence

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.