forming difference equation there is a square with $60$ equal blocks.
If a mosquito(bug)is set to fly starting at block $1$, it is equally likely to fly to other blocks. what is the probability after $n$ flies, the mosquito is at the $60\text{th}$ block (diagonally opposite to block $1$). form the difference equation.
I am trying to use law of probability to solve it but cannot form the equation.can anybody here can help. Thank you.
 A: We assume that the fly can be at any one of $60$ positions, and that in any flight, it chooses at random one of the $59$ other positions to go to. 
Say it starts at position $A$. We want to find the probability that after $n$ flights it ends up at $B$, where $B$ is a particular position other than $A$. 
Let this probability be $p_n$. We have $p_0=0$. The probability that after $n$ flights it ends up at $B$ is the probability it is at a position other than $B$ after $n-1$ flights, and goes to $B$ on the $n$-th. Thus
$$p_n=(1-p_{n-1})\cdot\frac{1}{59}.\tag{1}$$
There are various methods to solve linear recurrences. One way is to look first at the *homogeneous recurrence 
$$p_n=-\frac{p_{n-1}}{59},$$
which has the general solution $p_n=a\left(-\frac{1}{59}\right)^n$ for some constant $a$. Then we look for a particular solution of (1). Here we look for a constant solution $b$. So we want 
$b=(1-b)\cdot\frac{1}{59}$, which gives $b=\frac{1}{60}$. The general solution of the recurrence (1) is then 
$$p_n=a\left(-\frac{1}{59}\right)^n.$$
Choose $a$ to meet the initial condition $p_0=0$. We get $a=-\frac{1}{60}$, and therefore 
$$p_n=\frac{1}{60}\left(1-\frac{(-1)^n}{59^n}\right).$$
Remark: There are many other approaches to the solution.  
