Sum of digits

first of all...i think the 4th term will have 15 zeroes and not 13...

so the product is (2x10²+2)(2x10⁴+2)(2x10⁸+2)(2x10¹⁶+2)(2x10³²+2)

note that if u try and multiply the first two....
it turns out to be like this...

4x10⁶+4x10⁴+4x10²+4
which is nothing but 4040404
this pattern will remain with 8 and the product will be 808080808080808
and the next product will be 16161616161616161616161616161616 and the next consisting of all 32s....
so answer is 32(3+2)=160....

oops. yes ketan, it will be 15 0s r of the order(2ⁿ -1) but i m nt sure of ur ans.