1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Sequence formula help

  1. Oct 7, 2008 #1
    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. jcsd
  3. Oct 7, 2008 #2


    User Avatar
    Science Advisor

    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.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Discussions: Sequence formula help