I am looking to solve the following equation:
$$x = \sum_{i=1}^{N} r^i$$
I recognize that this is an order $N$ polynomial, which makes finding solutions hard for high $N$ (think of order $30$). Is there a way of solving a problem like this for $r$?
Context
I am trying to deduce the interest $r$ from a $N$-year investment. Because of tax deductions etc I can only work with a (constant) yearly investment $B$ and the total sum at the end $T$.
The payoff $T$ could be found as follows:
$$T = \sum_{i=1}^N Br^i = B\sum_{i=1}^nr^i$$
Factoring out $B$ gets us:
$$ \frac{T}{B} = x = \sum_{i=1}^Nr^i$$
Which I'm trying to solve.
So I am looking at positive real values for both $x$ and $r$.
Because I'm working with interest rates, $r$ will be close to 1, so $x \sim N$.
As you note, this is a fairly large polynomial in $r$, and so $r$ cannot be solved for in general (Abel-Ruffini theorem). We may, however, deduce some approximations.
For large $x$, we have
$$x\approx r^N\implies r\approx\sqrt[N]x$$
For small $x$, we have
$$x\approx r+r^2\implies r\approx\frac{-1\pm\sqrt{4x+1}}2$$
If $N$ is even, then we should consider both signs, if $N$ is odd, only consider the positive solution.
The full series expansion may be given by Lagrange inversion theorem,
$$r(x)=x+\sum_{n=2}^\infty\frac{x^n}{n!}\sum_{k=1}^{n-1}(-1)^kn^{(k)}a_{n,k}\\a_{n,k}=\begin{cases}B_{n-1,k}\left(\frac12,\frac13,\dots,\frac1{n-k+1}\right),&n-k<N\\B_{n-1,k}\big(\underbrace{\frac12,\frac13,\dots,\frac1N,0,\dots,0}_{n-k}\big),&n-k\ge N\end{cases}$$
where $B_{n,k}$ is a Bell polynomial.