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

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mathteacher1729
 4 years ago
Best ResponseYou've already chosen the best response.1There is a beautiful explanation of this without words here: http://books.google.com/books?id=cyyhZrSffcC&lpg=PA67&dq=proofs%20without%20words%20sum%20of%20cubes&pg=PA85#v=onepage&q&f=false

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I cant look at that and understand it, it means nothing to me.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Oh! 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
 4 years ago
Best ResponseYou've already chosen the best response.1the 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
 4 years ago
Best ResponseYou've already chosen the best response.0So it would be 496^2?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0So say it was the first 31 fourthpowers, it would be 496^3?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0But could you help me with my theory that I was trying to establish?

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1Look 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
 4 years ago
Best ResponseYou've already chosen the best response.1oh look, I just found this! http://www.mathpages.com/home/kmath279/kmath279.htm so I guess there is a pattern after all :D

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Okay, 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
 4 years ago
Best ResponseYou've already chosen the best response.0oh, great! I'll take a look at it.

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1I wish I could generalize this stuff, but I'm not that smart. Looks like somebody was though. Enjoy, I'm reading too.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I can't figure that out. I don't understand what it means. :( Now I'm frustrated...

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1The 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
 4 years ago
Best ResponseYou've already chosen the best response.1typo, first series should be\[\sum_{n=1}^{5}n=1+2+3+4+5\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0It's too small for me to read anyways, but thanks for trying.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I couldnt even begin to understand that link.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0the 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
 4 years ago
Best ResponseYou've already chosen the best response.0there is a way of deriving each of them from the previous one, but the algebra gets increasingly more painful
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