# Proof of a Sequence

1. Aug 23, 2004

Hello all

I encountered the following problem in my textbook

Prove (1^3 + 2^3 + ... + n^3 ) = ( 1 + 2 +... + n) ^2.

Here is my solution, however I become stuck on the final part of the problem.

Let S1 = 1/2(n+1)

Let S2 = 1/6n(n+1)(2n+1)

S3 = (1^3 + 2^3 + ... + n^3)

(1^3 + 2^3 + ... + n^3 ) = (v+1)^4 - v^4 = (v^4 + 4v^3 + 6v^2 + 4v + 1) - v^4

= 4v^3 + 6v^2 + 4v + 1.

Substituting v = 0 , 1 , 2, ... , n into the equation and adding we get

4(0)^3 + 6(0)^2 + 4(0) + 1​
4(1)^3 + 6(1)^2 + 4(1) + 1​
4(2)^3 + 6(2)^2 + 4(2) +1​
4(n)^3 + 6(n)^2 + 4(n) + 1​

= 4S3 + 6S2 + 4S1 + n + 1

(n+1) ^4 = 4S3 + 6S2 + 4S1 + n + 1

Since we are solving for S3, we collect terms and get

4S3 = (n+1) ^4 - 6S2 - 4S1 - n - 1

= (n+1) ([n+1]^3 - 1- n(2n+1) - 2n)

= (n+1) (n^3 + 3n^2 +3n + 1) - (1 + n(2n+1) +2n)

= (n+1) (n^3 + 3n^2 +3n +1 ) - (1 + 2n^2 +n +2n)

= (n+1) (n^3 + n^2 )

S3 = ¼ (n+1)(n^3 + n^2)

Here is where I become stuck.

How does S3 = ( 1 + 2 +... + n) ^2?

Any help would be greatly appreciated.

Thanks

Also do you know where I can get LaTex?

Last edited: Aug 23, 2004
2. Aug 23, 2004

### arildno

$$S3=\frac{n^{2}(n+1)^{2}}{2^{2}}=(\frac{n(n+1)}{2})^{2}=(\sum_{i=1}^{n}i)^{2}$$

Hope that helps..

3. Aug 23, 2004

### gravenewworld

Have you tried proof through induction? Thats the first thing that always comes to my mind when I see these types of problems.

4. Aug 23, 2004

Thanks a lot arildno. That really helped me!

5. Aug 23, 2004

### rgoudie

This exact same problem occurred in one of my final exams. The question required a poof by induction.

-Ray.

6. Aug 23, 2004

### rgoudie

Are you asking how to use LaTeX in these forums? One fast way to obtain the reference PDF is by clicking directly on arildno's formatted text. This will open a window that will contain a link.

-Ray.

7. Aug 24, 2004

how would i use the formula for 1^2 + 2^2 +... +n^2 = 1/6n(n+1)(2n+1) to find a formula for

1^2 + 3^2 +... (2n+1)^2. Would i use the same method i used in my other problem?

Any help would be appreciated.

Thanks!

8. Aug 24, 2004

### HallsofIvy

Staff Emeritus
1^2 + 3^2 +... (2n+1)^2 (sum of squares of odd numbers) is 1^2 + 2^2 +... +n^2- 2^2+ 4^2+ 6^2+ ... (sum of all squares minus sum of square of even numbers) and, of course, 2^2+ 4^2+ 6^2+... is 4(1^2+ 2^2+ 3^2+...).

Be careful about the limits! In order to find 1+ 3^2+ 5^2+ 7^2 (i.e. n= 3 since 7= 2(3)+1 you would find 1^2+ 2^2+ 3^2+ 4^2+ 6^2+ 7^2+ 8^2 - (2^2+ 4^2+ 6^2+ 8^2)
= 1^2+ 2^2+ 3^2+ 6^2+ 7^2+ 8^2- 4(1^2+ 2^2+ 3^2+ 4^2). You would then use the formula for sums of square with n= 8 and n= 4. In general, if we let S(n) mean "the sum of squares up to n^2" (the original formula), the sum of odds up to 2n+1 would be S(2n+2)- 4S(n+1).

9. Aug 24, 2004

Thanks a lot. That really helped!

10. Aug 24, 2004