Solving Coin Toss Problem If a coin is tossed 3 times,there are possible 8 outcomes.
HHH HHT HTH HTT THH THT TTH TTT
In the above experiment we see 1 sequnce has 3 consecutive H, 3 sequence has 2 consecutive H and 7 sequence has at least single H.
Suppose a coin is tossed n times.How many sequence we will get which contains a consequnce of H of length at least k?.How to solve this problem using recursive relation?

The number of binary strings of length $n$ with a sequence of ones of length at least 1 is, of course, $2^n-1$. Let's write $f(n,1)=2^n-1$. 
The number of binary strings of length $n$ with a sequence of ones of length at least 2 is tabulated at https://oeis.org/A008466 and the formula given is $f(n,2)=2^n-F_2(n+2)$. I'm writing $F_2(m)$ for the $m$th Fibonacci number. 
The number of binary strings of length $n$ with a sequence of ones of length at least 3 is tabulated at https://oeis.org/A050231 and the formula given is $f(n,3)=2^n-F_3(n+3)$. I'm writing $F_3(m)$ for the $m$th "tribonacci" number. 
The number of binary strings of length $n$ with a sequence of ones of length at least 4 is tabulated at https://oeis.org/A050232 and the formula given is $f(n,4)=2^n-F_4(n+4)$. I'm writing $F_4(m)$ for the $m$th "tetranacci" number. 
And the number of binary strings of length $n$ with a sequence of ones of length at least 5 is tabulated at https://oeis.org/A050233 and the formula given is $f(n,3)=2^{n+1}-F_5(n+6)$ [but I suspect it's supposed to be $f(n,5)=2^n-F_5(n+5)$]. I'm writing $F_5(m)$ for the $m$th "pentanacci" number. 
So I would hazard a guess that the number of binary strings of length $n$ with a sequence of ones of length at least $k$ is given by $f(n,k)=2^n-F_k(n+k)$ where $F_k(m)$ is the $k$th "$m$-nacci" number. 
