Given that s_n denotes the number of n-bit strings that do not contain the pattern 00, what is the recurrence relation?

There are two such possible strings, 0 and 1, so .

1 can be appended to either string, while 0 can only be appended to one, so .

For an -string to end with a 0, the bit in the th position needs to be a 1 - by definition, there are such possible strings.

If an -string already ends in a 0, we can drop this last bit to get an -string that ends in a 1 (we know this occurs because 00 is not allowed) - by definition there are such strings - and append 10 to it instead, giving an -string that ends in a 0.

So, the number of bit strings not containing consecutive 0s is given recursively by

Im sorry i cant answer it : ( im so sorry

Maybe fx = lxl because of the way the x is positioned is that what he