Best way to solve for $(1+a)^{-n}=1-\frac{I}{C}a$

Question

Best way to solve for $a$ for an equation that reads:

$$ C\cdot\frac{1-(1+a)^{-n}}{a}=I; n\geq 1. $$

This can be rewritten into:

$$ (1+a)^{-n}=1-\frac{I}{C}a. $$

One way to do it maybe is set $b=a+1$ and $X=\frac{I}{C}$, which would give:

$$ \frac{1}{b^n}=1-X(b-1). $$

Is there a meaningful solution to find $a$?

Answer 1

You got a polynomial equation in $b$ of degree "$n+1$"

$Xb^{n+1}-(X+1)b^n+1=0$

Sadly, we don't have any closed form solution for $b$ in terms of $n$.

Answer 2

for $n=1$ we get $$a=0$$ or $$a=\frac{C-J}{J}$$ for $n=2$ we get $$a=0,a=\frac{C-2J+\sqrt{C^2+4CJ}}{J}, a=\frac{1}{2}\frac{C-2J-\sqrt{C^2+4CJ}}{J}$$ for $n=3$ the solution looks complicated