n cubed equals ...

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

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.

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!

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

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.

It was my attempt to come up with a more kinaesthetic answer but it is a bit tricky without good graphics or unit cubes!