Please turn on javascript in your browser to play chess.
Posers and Puzzles
RHP Arms
Joined 09 Jun '07 Moves 48793 Originally posted by udayak
Here is another proof:
n^3 = n * n^2
Sum of an arithmetic series is:
S = n * m, where n = number of terms, and m = mean of the series.
In our case, in the series of consecutive odd numbers,
if m = n^2, it will satisfy.
For the series (of consecutive odd numbers), interval d=2.
Also,
Mean m = [2*a + (n-1)*d]/2 where a = first term, and ...[text shortened]... y the equation.
I think, this is better than the first one. No formulae used.
+Udaya Yes I like this better. I think there are many ways of proving this; the outcome is quite neat I think!
Joined 10 Aug '07 Moves 32581 Deriver,
I do get it. You take a cube, and divide it into so many squares as the cubic root, and try to build terms of the arithmetic sequence.
But, do you not construct the same arithmetic sequence with the trick.
I was also trying to see what the different end terms would look like geometrically. Dumb! That would not make sense.
+Udaya
Edit: Did not look at the last reply. Included the name "deriver", as it is his reply.
Joined 10 Aug '07 Moves 32581 Originally posted by wolfgang59
Yes I like this better. I think there are many ways of proving this; the outcome is quite neat I think! I never knew that there was such a relation.
Another interesting thing that came out of the first proof is that:
For any n, n^3 can always be expressed as a difference between two squares.
1^3 = 1^2 - 0^2
2^3 = 3^2 - 1^2
3^3 = 6^2 - 3^2
4^3 = 10^2 - 6^2
5^3 = 15^2 - 10^2
Or in general, n^3 = S(n)^2 - S(n-1)^2
where S(n) = 1 + 2+ ... + n
+Udaya
RHP Arms
Joined 09 Jun '07 Moves 48793 Originally posted by udayak
I never knew that there was such a relation.
Another interesting thing that came out of the first proof is that:
For any n, n^3 can always be expressed as a difference between two squares.
1^3 = 1^2 - 0^2
2^3 = 3^2 - 1^2
3^3 = 6^2 - 3^2
4^3 = 10^2 - 6^2
5^3 = 15^2 - 10^2
Or in general, n^3 = S(n)^2 - S(n-1)^2
where S(n) = 1 + 2+ ... + n
+Udaya .. and the difference of the squares of two consecutive triangular numbers is a cube!
Joined 16 Jul '08 Moves 1693 Originally posted by geepamoogle
1^3 = 1 = 1
2^3 = 8 = 3 + 5
3^3 = 27 = 7 + 9 + 11
4^3 = 64 = 13 + 15 + 17 + 19
5^3 = 125 = 21 + 23 + 25 + 27 + 29
Pattern continues.
Prove that the sum of the first n cubes is the square of the sum of the first n numbers. well , in each case , do you see that you get n pairs of the actual square n ?
n * n square = cubbed n
1^3 = 1 = 1 (n=1)
2^3 = 8 = 3 + 5 (3+5)/4 = 2 rest 0 (one pair or two's)
3^3 = 27 = 7 + 9 + 11 (7+11)/2 = 2*9 + 9 = 3*9 , where n=3
4^3 = 64 = 13 + 15 + 17 + 19 (13+19)/2+(15+17)/2 = 2*16+2*16 = 4*16 , where n = 4
actually the quote gave me the hint 😉
Joined 28 Mar '07 Moves 957 It was my attempt to come up with a more kinaesthetic answer but it is a bit tricky without good graphics or unit cubes! I didn't follow all the maths, but your explanation made perfect sense to me. I think you nailed it, even without props.
Phil.
Cookies help us deliver our Services. By using our Services or clicking I agree, you agree to our use of cookies. Learn More . I Agree