anonymous
  • anonymous
How would I find the sum of the first 31 cubes?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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mathteacher1729
  • 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
anonymous
  • anonymous
I cant look at that and understand it, it means nothing to me.
UnkleRhaukus
  • UnkleRhaukus
that's beautiful

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anonymous
  • 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?
TuringTest
  • TuringTest
the formula is printed very clearly on mathteacher's link\[\large1^3+2^3+3^3+\dots+n^3=(1+2+3+\dots+n)^2\]
anonymous
  • anonymous
So it would be 496^2?
anonymous
  • anonymous
So say it was the first 31 fourth-powers, it would be 496^3?
anonymous
  • anonymous
But could you help me with my theory that I was trying to establish?
TuringTest
  • 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.
TuringTest
  • 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
anonymous
  • 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?
anonymous
  • anonymous
oh, great! I'll take a look at it.
TuringTest
  • TuringTest
I wish I could generalize this stuff, but I'm not that smart. Looks like somebody was though. Enjoy, I'm reading too.
anonymous
  • anonymous
I can't figure that out. I don't understand what it means. :( Now I'm frustrated...
TuringTest
  • 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.
TuringTest
  • TuringTest
typo, first series should be\[\sum_{n=1}^{5}n=1+2+3+4+5\]
anonymous
  • anonymous
It's too small for me to read anyways, but thanks for trying.
anonymous
  • anonymous
I couldnt even begin to understand that link.
anonymous
  • 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
anonymous
  • anonymous
yes.
anonymous
  • anonymous
there is a way of deriving each of them from the previous one, but the algebra gets increasingly more painful

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