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anonymous

  • 4 years ago

How would I find the sum of the first 31 cubes?

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  1. mathteacher1729
    • 4 years ago
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    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
    • 4 years ago
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    I cant look at that and understand it, it means nothing to me.

  3. UnkleRhaukus
    • 4 years ago
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    that's beautiful

  4. anonymous
    • 4 years ago
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    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
    • 4 years ago
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    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
    • 4 years ago
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    So it would be 496^2?

  7. anonymous
    • 4 years ago
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    So say it was the first 31 fourth-powers, it would be 496^3?

  8. anonymous
    • 4 years ago
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    But could you help me with my theory that I was trying to establish?

  9. TuringTest
    • 4 years ago
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    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
    • 4 years ago
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    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
    • 4 years ago
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    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
    • 4 years ago
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    oh, great! I'll take a look at it.

  13. TuringTest
    • 4 years ago
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    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
    • 4 years ago
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    I can't figure that out. I don't understand what it means. :( Now I'm frustrated...

  15. TuringTest
    • 4 years ago
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    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
    • 4 years ago
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    typo, first series should be\[\sum_{n=1}^{5}n=1+2+3+4+5\]

  17. anonymous
    • 4 years ago
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    It's too small for me to read anyways, but thanks for trying.

  18. anonymous
    • 4 years ago
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    I couldnt even begin to understand that link.

  19. anonymous
    • 4 years ago
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    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
    • 4 years ago
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    yes.

  21. anonymous
    • 4 years ago
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    there is a way of deriving each of them from the previous one, but the algebra gets increasingly more painful

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