Is this for a loan repayment? If so, consider the following, which can easily be changed to an annual payment:
Let $P$ = monthly payment, $A$ = amount borrowed, $i$ = monthly interest rate, and $n$ = the number of payments.
Also, let $D_n$ be the debt amount after payment $n$.
Consider the recursion:
(1) [math]D_{n}=(1+i)D_{n-1}-P[/math]
(2) [math]D_{n+1}=(1+i)D_{n}-P[/math]
Subtracting (1) from (2) yields the homogeneous recursion:
[math]D_{n+1}=(2+i)D_{n}-(1+i)D_{n-1}[/math]
whose associated auxiliary equation is:
[math]r^2-(2+i)r+(1+i)=0[/math]
[math](r-(1+i))(r-1)=0[/math]
Thus, the closed-form for our recursion is:
[math]D_n=k_1(1+i)^n+k_2[/math]
Using initial values, we may determine the coefficients $k_i$:
[math]D_0=k_1+k_2=A[/math]
[math]D_1=k_1(1+i)+k_2=(1+i)A-P[/math]
Solving this system, we find:
[math]k_1=\frac{Ai-P}{i},\,k_2=\frac{P}{i}[/math] and so we have:
[math]D_n=\left(\frac{Ai-P}{i} \right)(1+i)^n+\left(\frac{P}{i} \right)=\frac{(Ai-P)(1+i)^n+P}{i}[/math]
Now, equating this to zero, we can solve for $P$:
[math]\frac{(Ai-P)(1+i)^n+P}{i}=0[/math]
[math](Ai-P)(1+i)^n+P=0[/math]
[math](P-Ai)(1+i)^n=P[/math]
[math]P\left((1+i)^n-1 \right)=Ai(1+i)^n[/math]
[math]P=\frac{Ai(1+i)^n}{(1+i)^n-1}[/math]
[math]P=\frac{Ai}{1-(1+i)^{-n}}[/math]