The sum $S_n$ of the first $n$ terms of a geometric series is given by $S_n=\frac{a(r^n-1)}{r-1}$, where $a$ is the first term and $r$ is the common ratio.
How do you solve an equation with this following form (solve for $x$):
$$y=\frac{x^n-1}{x-1} + c$$ $n$ and $c$ are both real numbers. The above equation is another form of the geometric series formula, since $a=1$, $r=x$, and $S_n=-c$. Since it is similar to the sum of a geometric series formula, my question can also be interpreted as, how to find $r$ given $a$, $n$, and $S_n$ of a geometric series.
Methods like long division might not work since the values of $n$ can be very large. Is there an inverse form of the above equation?
Example question: Solve for $x$:


The equation is

$10=\frac{x^{36}-1}{x-1}\qquad |\cdot (x-1)$



This is a polynomial equation of degree 36. This equation can not solved algebraically. $x=1$ could be a solution, but the first equation shows that $x\neq 1$ (denominator). You have to apply an approximation method, like the Newton-Raphson method, to solve the equation.

  • $\begingroup$ Do I have to use Newton's method? Is there an inverse of the equation? $\endgroup$ – MattMath Jan 15 '17 at 22:30
  • 1
    $\begingroup$ You have to use an approximation method. The newton raphson method is one of them (my favorite). But there are others like Bisection method. No, there exists no inverse of this equation. $\endgroup$ – callculus Jan 15 '17 at 22:33

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