Ask Me Help Desk - Solve in terms of n

OK. Simple. (once I've shown you ;) )

$f=c\left({\left(1+r\right)}^n + \frac{{\left(1+r\right)}^{n+1}-\left(1+r\right)}{r}\right) \\ \text{divide by c} \\ \frac f c={\left(1+r\right)}^n + \frac{{\left(1+r\right)}^{n+1}-\left(1+r\right)}{r} \\ \text{multiply by r} \\ \frac {fr} c=r{\left(1+r\right)}^n + {\left(1+r\right)}^{n+1}-\left(1+r\right) \\ \text{expand the middle term} \\ \frac {fr} c=r{\left(1+r\right)}^n + {\left(1+r\right)}^n\left(1+r\right)-\left(1+r\right) \\ \text{collect the } \left(1+r\right)^n \text{terms}\\ \frac {fr} c=\left(r+\left(1+r\right)\right){\left(1+r\right) }^n -\left(1+r\right) \\ \text{simplify } r+(1+r) \text{and add} (1+r) \\ \frac {fr} c + \left(1+r\right) = \left( 2r+1\right){\left(1+r\right)}^n \\ \text{divide by } 2r+1 \\ \frac {\frac {fr} c + \left(1+r\right)} {\left( 2r+1\right)} = {\left(1+r\right)}^n \\ \text{simplify the left by multiplying by } \frac c c \\ \frac {fr + c + cr} {c\left( 2r+1\right)} = {\left(1+r\right)}^n \\ \text{take the base } \left( 1+r\right) \text{logarithm} \\ \log_{1+r}\left(\frac {fr + c + cr} {c\left( 2r+1\right)}\right) = \frac {\log\left(\frac {fr + c + cr} {c\left( 2r+1\right)}\right)} {\log(1+r)} = n \\ $

got the same answer as galactus' TI92 so I think it looks good.

(note: also confirmed via Maxima)