Solve for "R" in classic annuity formula $$p=\frac{c}r*(1-\frac1{(r+1)^t} )$$
I'm stuck.  I'm building an excel model where I will be able to put in inputs for p, c, and t... but it will need to solve for "r."
For some reason, I can't figure out the algebra.  
Can you help me solve for "r"?   
 A: You could try Newton's method to solve for $r.$ Take,
$$F(r) = p-\frac{c}r\cdot \left(1-\frac1{(r+1)^t}\right)$$
then your goal is to find $r$ such that $F(r)=0.$ For this you start with some choice $r_0$ and then use the following recursive definition:
$$r_{n+1}= r_n - \frac{F(r_n)}{F'(r_n)}.$$
This will converge to the root of $F.$
A: I prefer to add a second answer.
Being just fascinated by David W. Cantrell's approximation
$$\color{green}{r\simeq \left(\left(1+\frac{c}{p}\right)^{\frac{1}{q}}-1\right)^q-1}\qquad \text{where} \qquad \color{green}{q=\log_2\left(1+\frac 1t \right)}$$ totally inspired by it, I tried something in the same spirit
$$\frac{c}r \left(1-\frac1{(r+1)^t}\right)=p\implies 1+\frac{c}{p}=1+\frac{r}{1-(r+1)^{-t}}$$ Taking logarithms of both sides
$$\log \left(1+\frac{c}{p}\right)=\log \left(1+\frac{r}{1-(r+1)^{-t}}\right)$$ Now, expanding the rhs as a Taylor series at $r=0$
$$\log \left(1+\frac{c}{p}\right)=\log \left(1+\frac{1}{t}\right)+\frac{r}{2}+\frac{2t-5}{24} r^2 +O\left(r^3\right)$$
Neglecting the second order term, we the obtain the first approximation
$$\color{blue}{r_1 =2 \log \left(\frac{t \,(c+p)}{p\, (t+1)}\right)}$$
Using the complete expansion to $O\left(r^3\right)$, we then have the second approximation
$$\color{blue}{r_2=\frac{\sqrt{1+4\, \alpha\,  r_1}-1}{2 \alpha }}\qquad \text{where}\qquad \color{blue}{\alpha=\frac {2t-5}{12}}$$
We could even avoid quadratic equations building the simplest Padé approximant instead of the Taylor series. This gives
$$\log \left(1+\frac{c}{p}\right)=\frac{\log \left(1+\frac{1}{t}\right)+\frac{1}{12} \left((5-2 t) \log
   \left(1+\frac{1}{t}\right)+6\right) r} {1+ \frac{5-2t}{12} r }\implies  
\color{blue}{r_3=\frac{12\,r_1}{12+(2t-5)\,r_1}}$$
Applied to the worked example, this would give 
$$r_1=2 \log \left(\frac{606}{605}\right)\approx 0.00330306$$
$$r_2=\frac{2}{235} \left(\sqrt{9+1410 \log \left(\frac{606}{605}\right)}-3\right)\approx 0.00311325$$
$$r_3=\frac{12 \log \left(\frac{606}{605}\right)}{6+235 \log
   \left(\frac{606}{605}\right)}\approx 0.00310238$$
 while the exact solution is $$r=0.00311418$$
