How can S3 be equal to the square of the sum of 1 to n?

[courtrigrad]

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?

 

[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..

 

[gravenewworld]

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

 

[courtrigrad]

Thanks a lot arildno. That really helped me!

 

[rgoudie]

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

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

-Ray.

 

[rgoudie]

Also do you know where I can get LaTex?

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.

 

[courtrigrad]

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!

 

[HallsofIvy]

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).

 

[courtrigrad]

Thanks a lot. That really helped!

 

[courtrigrad]

HallsofIvy, is that just a typo or did you make a mistake in counting 5^2 and 4^2?

 

[HallsofIvy]

Yeah, I dropped 5^2 in one sum and both 5^2 and 4^2 in another. Thanks for catching that.