Look at this

1 = 1 = 1^3

3 + 5 = 8 = 2^3

7 + 9 + 11 = 27 = 3^3

13 + 15 + 17 + 19 = 64 = 4^3

...

nice, isnt'it?

The proof is not that easy

Bye

Printable View

- Apr 12th 2005, 07:29 AMtheprofOdd sums give cubes
Look at this

1 = 1 = 1^3

3 + 5 = 8 = 2^3

7 + 9 + 11 = 27 = 3^3

13 + 15 + 17 + 19 = 64 = 4^3

...

nice, isnt'it?

The proof is not that easy

Bye - Apr 18th 2005, 03:47 PMstringmana simple solution
The solution is easy.

The sum is always of the type of q odd integer starting from 2 p+1 ending with 2(p+q)+1

S_p,q=\Sum_k=p^p+q-1 (2k+1) = q (2p +q)

At the first set we sum 1 term so q=1 and p=0, at the second step we sum 2 terms so q=2 and p=1 etc

We just have to find the starting integer p at step q.

But at set q we have skipped 1 + 2 + ... + q-1 = q(q-1)/2 odd integers therefore

p = q (q-1)/2

With this value of p

S_p,q = q^3

QED