Define the sequence of positive integers a_n recursively by a_1=3 and a_n=3(a_n- 1) for all n> =2. Determine the last two digits of a_{2007}.
The GP is as follows:
3, 9, 27, 81, 243,................a(2007) ==3^2007 mod 10^10 ==9,532,282,187 - these are the last 10 digits of a(2007)