Added by ahish on May 04, 2001 at 12:48:07:
the sum of squares of first 'n' natural no`s-1^2+2^2+3^2+.....+n^2=n*(n+1)*(2n+1)/6
the sum of cubes of first 'n' natural no`s-1^3+2^3+3^3+4^3+......+n^3=(n^2 * (n+1)^2)/4
happy calculating!!!!