Re: combinatorial argument
Perhaps it is easier to show the following :
1^2+2^2+3^2+...+n^2 = 2*(n+1)c3+(n+1)c2 for n>=2
1^3+2^3+3^3+...+n^3 = 6*(n+1)c4 + 6*(n+1)c3 + (n+1)c2 for n>=3
Notice that the rhs of the second equation reduces to [(n+1)c2]^2 . Pretty interesting , nonetheless .
|