How do I find the closed form of a recurrence relation? I'm stuck on how to find closed forms of recurrence relations. My current problem is: 
An employee joins a company in 1999 with a starting salary of $50,000. Every year this employee receives a raise of 1,000 plus 5% of the salary of the previous year.
The basic setup I have for the relation is:
A0 = 50,000
A1 = 53,500
A2 = 57,175
An+1 = 1.05An + 1000
My problem is on finding an explicit formula for the salary of the employee n years after 1999. I believe this is called the closed form. I'm stuck! Can anyone help?
 A: For me, it's easier to establish a pattern from general values.  Let $P_0$ be the initial salary, $P_n$ be the salary after the $n$th year, $D = 1000$ be the fixed raise, and $r=0.05$ be the raise rate.
You can then write down the first few values
$$P_1 = (1+r) P_0 + D$$
$$P_2 = (1+r) P_1 + D = (1+r)^2 P_0 + [1+(1+r)]D$$
$$P_3 = (1+r) P_2 + D = (1+r)^3 P_0 + [1+(1+r)+(1+r)^2]D$$
I hope you can see that
$$P_n = (1+r)^n P_0 + \left ( \sum_{k=0}^{n-1} (1+r)^k \right ) D$$
Evaluating the geometric sum, we get
$$P_n = (1+r)^n P_0 +  \frac{(1+r)^n - 1}{r} D$$
If the employee started in 1999, then in 2013, (s)he is making $\$118,595$.
A: You have the right recurrence set up.
$$A_{n+1} = 1.05 A_n + 1000 \implies \left(A_{n+1} + x \right) = 1.05 \left(A_n + x\right)$$
where we need $1.05x -x = 1000 \implies x = \dfrac{1000}{0.05} = 2 \times 10^4$. Hence, if we let $B_ n =A_n + 2 \times 10^4$, we get that
$$B_{n+1} = 1.05 B_n$$ I am sure you can take it from here to get $B_n$ and thereby $A_n$.
A: Hint: $$A_2=1000 + 1.05 A_1 = 1000+ 1.05(1000 + 1.05\cdot 50000) = 1000 + 1.05\cdot 1000 + 1.05^2\cdot 50000$$
Then $$A_3= 1000 + 1.05A_2 = 1000 + 1.05\cdot 1000 + 1.05^2\cdot 1000 + 1.05^3\cdot 50000$$
Start looking for a pattern here first.
