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- Jul 27th 2010, 02:19 PMfxs12Sum of cubes.
- Jul 27th 2010, 02:22 PMmr fantastic
- Jul 27th 2010, 02:43 PMAlso sprach Zarathustra
1^3+2^3+3^3+...+n^3=(1^2+2^2+3^2+...+n^2)^2={(n(n+ 1)/2}^2

Someway to prove the formula:

An^4+Bn^3+Cn^2+Dn+E=S_n

put:

n=1,2,3,4

and solve the system of 4 equations, find A,B,C,D,E. - Jul 27th 2010, 02:52 PMSoroban
Hello, fxs12!

How about an inductive proof?

Quote:

. . . True.

. . .[1]

. . . . . . . . . . . .

Therefore, [1] becomes: .

The inductive proof is complete.