Explanation :
Let a_{n} be the number of n-bit strings that do not contain two consecutive 1's. we wish to develop a recurrence relation for a_{n}.
Consider 1 bit strings 0, 1
So $ a_1=2 $
Consider 2 bit strings 00, 01, 10, 11
Out of minimum only 00, 01, 10 do not contain two consecutive 1's.
Consider 3 bit strings
Out of minimum six strings only 000, 001, 010, 100 and 101 five strings satisfy do not contain two consecutive 1's.
So $a_3=5$. There numbers $ a_1,\;a_2,\;a_3 $ satisfy clearly only (b) $ a_n=a_{n-1}+a_{n-2} $ correct.