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