With every cut you should cut through every previous cut. As long as you keep doing this you will always be able to do so again your next cut. The number of additional pieces formed on cut x will always be x. Everyone should know that sum of all integers from 1 to n is n*(n+1)/2. So the sum of additional pieces cut will be n*(n+1)/2. Finally add 1 because we start with 1 piece: n*(n+1)/2 + 1.

Michael Shackleford, A.S.A., 11/10/1998