# Proving the sum of squared odd numbers

1. Jun 28, 2012

### Duderonimous

1. The problem statement, all variables and given/known data
A problem out of "What is Mathematics" by Courant and Robbins

Prove that

1$^{2}$+3$^{2}$+...+(2n+1)$^{2}$ =

$\frac{(n+1)(2n+1)(2n+3)}{3}$

2. Relevant equations

Prove this using

1$^{2}$+2$^{2}$+...+n$^{2}$ =

$\frac{n(n+1)(2n+1)}{6}$

3. The attempt at a solution

I tried seeing how the ratio of 1st, 2nd, and 3rd terms of each sequence change as one goes on in the sequence. Like 1/1 =1, 5/10 =.5, 14/35=.4, 30/84=.357142857143... I didn't get much insight there. I know I have to start with the second equation and divide or multiply it by something to get the first equation. Cant figure out that term would be. Any help would be great! Thanks.

2. Jun 28, 2012

### Staff: Mentor

start with the 1^2 + 2^2 + 3^2 ... which you know the sum formula for to

and subtract 2 * (1^2 + 2^2 + 3^2 ...) from it to get 1^2 + 3^2 + 5^2

do you see the solution now?

3. Jun 28, 2012

### Staff: Mentor

I don't see that looking at the ratios will get you anywhere.

You are given the formula for the sum of the squares of the first n integers. Use that formula to get the sum of the squares of the first (2n + 1) integers, and then subtract the sum of the squares of the first n even integers.

As a simple example, $1^2 + 2^2 + 3^2 + 4^2 + 5^2 = (1^2 + 3^2 + 5^2) + 2^2 + 4^2)$, so
$(1^2 + 3^2 + 5^2) = 1^2 + 2^2 + 3^2 + 4^2 + 5^2 - (2^2 + 4^2)$

That last part can be simplified to $2^2(1^2 + 2^2)$.

4. Jun 28, 2012

### tt2348

So. You have to find a formula for a sum over odd squared integers.
Start with $1+2^2+3^2....+(2n)^2+(2n+1)^2$ , which is going to be your sum of the odd squares plus 4*(1+2^2+3^2....+n^2). Then f(n)=1+2^2...+n^2 , you'll have f(2n+1)=x+4*f(n) (x being your odd sum).

Last edited: Jun 28, 2012
5. Jun 28, 2012

### Duderonimous

Ahh! Thank you so much both of you. It's amazing how the solution is pretty much smacking you in the face, but nonetheless I am numb.

6. Aug 14, 2012

### Duderonimous

f(n)=(12+32+...+(2n+1)2)+(22+42+...+(2n)2)

$\frac{n(n+1)(2n+1)}{6}$=(12+32+...+(2n+1)2)+4(12+22+...+n2)

$\frac{n(n+1)(2n+1)}{6}$=(12+32+...+(2n+1)2)+4($\frac{n(n+1)(2n+1)}{6}$)

(12+32+...+(2n+1)2)=$\frac{n(n+1)(2n+1)}{6}$-4($\frac{n(n+1)(2n+1)}{6}$)

(12+32+...+(2n+1)2)=-$\frac{n(n+1)(2n+1)}{2}$

I cant figure out how to derive $\frac{(n+1)(2n+1)(2n+3)}{3}$

Any help is appreciated!

7. Aug 14, 2012

### eumyang

Just because
$1^2 + 2^2 + ... + n^2 = \frac{n(n + 1)(2n + 1)}{6}$
doesn't mean that
$1^2 + 2^2 + ... + (2n + 1)^2 = \frac{n(n + 1)(2n + 1)}{6}$
... which looks like what you are saying. What you need to do is plug in "2n + 1" for "n" on the right side of the formula:
$$1^2 + 2^2 + ... + (2n + 1)^2 = \frac{(2n + 1)[(2n + 1) + 1][2(2n + 1) + 1]}{6}$$
Try it from here.

8. Aug 15, 2012

### Duderonimous

Thank you much.

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