Find the recurrence relation for the number of bit strings that contain the string $01$. Question:Find the recurrence relation for the number of bit strings that contain the string $01$.
Attempt:
Since $01$ can appear in a lot of places, I focused on instances without $01$ first.
Bit strings without $01$ in a bit string of length $n$. Let $XX\dots X$ be bit string without $01$:


*

*$XX\dots X0$

*$1XX\dots X$


Thus we know we acquire twice more bit string without $01$ from length of $n-1$ to $n$. Let $a_n$ be the count of bit strings without $01$ at length $n$, recurrence relation of this is the following:
$$a_n = 2a_{n-1}, a_2 = 1$$
The inverse of this is then the recurrence relation with $01$. Let $P_n$ be the recurrence relationship of the number of bit string with length $n$ with $01$, thus,
$$P_n = 2^n - a_n = 2^n - 2a_{n-1}, a_2 = 1$$
Since $P_n = 2^n - a_n$, then, $a_n = 2^n - P_n$, thus
$$P_n = 2^n - 2*(2^{n-1} - P_{n-1})$$
$$\iff P_n = 2^n - 2^n + 2P_{n-1}$$
$$P_n = 2P_{n-1}, P_2 = 1$$
Problem:
I've also played around with the n-combination to solve this. The format of a bit string without $01$ is the following,
$$1\dots10\dots0$$
Adding $0$ or $1$ to either side adds a $01$, thus we can only add either $1$ on the left and $0$ on the right. Another way to view this is,
$1^p0^q, p, q \geq 1$
The number of bit strings without $01$ is for a bit string of length $n$,
$${2 + n - 2 - 1\choose n-2} = \\
{n-1 \choose 1} = \\
n-1$$
The inverse of this is,
$$2^n - n+1$$
Which is the number of bit string with $01$ at length $n$.
If we solve the recurrence relation above for $P_n$,
$$P_3 = 2P_2 = 2$$
$$P_4 = 2*2\\
\vdots\\
P_n = 2^{n-2}$$
Which is inconsistent with the $n$-combination solution to solve this. Where did I get it wrong??
 A: 
Where did I get it wrong??

The number of bit-strings of length $n$ without 01 is: $n+1$. (Not $n-1$.)
You have the pattern $\underbrace{1\cdots 1}_{k}\underbrace{0\cdots 0}_{n-k}$, where the number of 1 bits, $k$, can range from $0$ to $n$.
Thus the number of $n$ length bit-strings with at least 1 substring of 01 is: $2^n-n-1$.


Thus we know we acquire twice more bit string without 01 from length of $n−1$  to $n$. 
Let $a_n$ be the count of bit strings without 01  at length n , recurrence relation of this is the following:
$$a_n =2a_{n−1} ,a_2 =1$$

You are over counting.  Adding 0 to the end of 110 gives the same result as adding 1 to the start of 100: namely 1100.  Just count the ways to add 0 or 1 to the end of the strings and remain valid.  You can viably add 0 to every valid $n-1$ length string; but there is only one viable target to which you can add 1 (the one with all 1s).
Also there are three string of length 2 which do not contain 01: namely 00, 10, 11.
$$a_n = a_{n-1}+1, a_2=3$$
Thus the valid 3-length strings are: 000, 100, 110, and 111.  $a_n=4$ as expected.
(Always test your logic on the simplest examples. )

The inverse of this is then the recurrence relation with 01 . Let $P_n$ be the recurrence relationship of the number of bit string with length $n$  with 01 , thus,

Applying the correct formula:
$$\begin{align}
P_n & = 2^n - a_{n}
\\ & = 2\cdot 2^{n-1} - (a_{n-1} + 1)
\\ & = P_{n-1} + 2^{n-1} - 1
\\[2ex]
P_2 & = 1 & 2^2-3
\\P_3 & = 1+4-1 = 4 & 2^3-4
\\P_4 & = 4+8-1 = 11 & 2^4 - 5
\\ \text{et cetera}
\end{align}$$
A: Every following "form" is a $n$ bit string.
Bit strings of the form ....$\_$ $\_$$0$ contributes $a_{n-1}$ to the total count.
Bit strings of the form ....$\_$ $\_$$01$ contributes $2^{n-2}$ to the total count.
Bit strings of the form ....$\_$ $\_$$011$ contributes $2^{n-3}$ to the total count.
$\vdots$
Bit strings of the form $011...1$  contributes $2^{n-n}$ to the total count.
Bit strings of the form $111...1$  contributes nothing to the total count.
All the bit strings of length $n$ have been considered.

Therefore, 
$a_{n}= a_{n-1}+2^{n-2}+2^{n-3}+...+2^{n-n}$ 
applying sum of G.P,we get,
$$a_{n}= a_{n-1}+2^{n-1}-1$$
A: I believe the problem with your recurrence relation is that you are overcounting the $a_n$.  
For example, the strings of length 3 without 01 are 111, 000, 100, and 110; 
but there are only 5 strings of length 4 without 01: 1111, 0000, 1000, 1100, and 1110.  
Notice that 1|110 gives the same result as 111|0, so the string
1110 is getting counted twice in your recurrence relation.

If we let $d_n$ be the number of bit strings of length n with 01, $e_n$ be the number of such bit strings ending with 0, and $f_n$  be the number of such strings ending in 1, then
$d_n=e_n+f_n$,  $e_n=d_{n-1}$, and $f_n=d_{n-1}+2^{n-2}-e_{n-1}$,
(since a string of length n containing 01 and ending in 1 either has 01 in the first n-1 bits, has any bits in the first n-2 places and a 0 in the n-1 spot, or satisfies both conditions).
This should yield a recurrence relation for $d_n$.
A: Case 1: starts with $1$ followed by strings of length $n-1$ containing $01$
Case 2: starts with $01$ followed by any string of length $n-2$, for which there are $2^{n-2}$ possibilities.
So we have $a_n = 2^{n-2} + a_{n-1}$ for $n\ge 2$.
