Proving the sum of squared odd numbers

[Duderonimous]

Homework Statement

A problem out of "What is Mathematics" by Courant and Robbins

Prove that

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

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

Homework Equations

Prove this using

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

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

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.

 

[jedishrfu]

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?

 

[Mark44]

Homework Statement

A problem out of "What is Mathematics" by Courant and Robbins

Prove that

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

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

Homework Equations

Prove this using

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

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

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.

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 $2^2(1^2 + 2^2)$.

 

[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).

 

[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.

 

[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 can't figure out how to derive $\frac{(n+1)(2n+1)(2n+3)}{3}$

Any help is appreciated!

 

[eumyang]

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)

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.

 

[Duderonimous]

Thank you much.