The number of terms in the expansion of (a3+1a3+1)100  is

(a3+1a3+1)100=1+C  100 1(a3+1a3)+C  100 2(a3+1a3)2+.....+C  100 100(a3+1a3)100 =a0+a1a3+a2a6+...+a100a300+b1a3+b2a6+...+b100a300  Where a0=  sum of all absolute terms =1+C  100 2.2+....  Similarly a1, a2, ....a100  and b1, b2, ....b100  are coefficients obtained after simplification. ∴  Total number of terms =1+100+100=201