You can solve it by formulating the recurrence-
Let T(N) be the recurrence relation for n length such strings
For n =1 we have nothing
For n=2 we have 00
For n=3 we have T(1)1 (Appending 1 to length n -1 strings which means Strings ending with 1 ) + 00(1) (Strings ending with 00)
For n=4 we have T(3)1 (Strings ending with 1) + (2^(2)-T(2))00 ( This means length n -2 strings which do not have consecutive zeroes.)
For n=5= we have T(4)1 + (2^3-T(3))00
So for n=n We have T(n)=T(n-1)+ (2^(n-2)) - T(n-2)