anonymous 4 years ago How would I find the sum of the first 31 cubes?

1. mathteacher1729

There is a beautiful explanation of this without words here: http://books.google.com/books?id=cyyhZr-SffcC&lpg=PA67&dq=proofs%20without%20words%20sum%20of%20cubes&pg=PA85#v=onepage&q&f=false

2. anonymous

I cant look at that and understand it, it means nothing to me.

3. UnkleRhaukus

that's beautiful

4. anonymous

Oh! Epic sauce look what I found! 1 8 27 64 125 7 19 37 61 12 18 24 - Multiples of 6! So something like (6x1 + 6x2 + 6x3 + 6x4...) which is 6(1 + 2 + 3 + 4... + n)! So with my theory the answer should be 6(1 + 2 +... + 31) so 6 x 496 is definitely not the answer :( way too small I think it has something to do with how for the first 5 cubes, i only had 3 multiples of 6, maybe multiply that by n + 2, so 33? That equals 98,208, which is still way too small. The smallest possible answer is 216,225. Help please?

5. TuringTest

the formula is printed very clearly on mathteacher's link$\large1^3+2^3+3^3+\dots+n^3=(1+2+3+\dots+n)^2$

6. anonymous

So it would be 496^2?

7. anonymous

So say it was the first 31 fourth-powers, it would be 496^3?

8. anonymous

But could you help me with my theory that I was trying to establish?

9. TuringTest

Look at the formula for the sum if the first n integers raise to power 1, 2, and 3 respectively:$\sum_{i=1}^{n}i=\frac{n(n+1)}{2}$$\sum_{i=1}^{n}i^2=\frac{n(n+1)(2n+1)}{6}$$\sum_{i=1}^{n}i^3=\frac{n^2(n+1)^2}{4}=(\sum_{i=1}^{n}i)^2$There is almost a pattern here, but not really. I don't know of any formulas for higher power series, let alone patterns among them all. I'm not sure such patterns are even fully known or understood.

10. TuringTest

oh look, I just found this! http://www.mathpages.com/home/kmath279/kmath279.htm so I guess there is a pattern after all :D

11. anonymous

Okay, I'm not sure what the greek letter means, but I think I get the gist of them. Thanks for your time. Could you help me out with my theory though?

12. anonymous

oh, great! I'll take a look at it.

13. TuringTest

I wish I could generalize this stuff, but I'm not that smart. Looks like somebody was though. Enjoy, I'm reading too.

14. anonymous

I can't figure that out. I don't understand what it means. :( Now I'm frustrated...

15. TuringTest

The greek letter is sigma. It just means add up all the numbers from the bottom to the top part$\sum_{i=n}^{5}n=1+2+3+4+5$for instance, or$\sum_{n=1}^{3}n^2=1^2+2^2+3^2$and yes, I never said that the pattern would be obvious. Good luck deciphering it.

16. TuringTest

typo, first series should be$\sum_{n=1}^{5}n=1+2+3+4+5$

17. anonymous

It's too small for me to read anyways, but thanks for trying.

18. anonymous

I couldnt even begin to understand that link.

19. anonymous

the sum of square is annoying, but the sum of cubes is actually nicer, since it is just the square of the sum of consecutive natural numbers

20. anonymous

yes.

21. anonymous

there is a way of deriving each of them from the previous one, but the algebra gets increasingly more painful