How to get sum of $\frac{1}{1+x^2}+\frac{1}{(1+x^2)^2}+...+\frac{1}{(1+x^2)^n}$ using mathematical induction Prehistory: I'm reading book. Because of exercises, reading process is going very slowly. Anyway, I want honestly complete all exercises.
Theme in the book is mathematical induction. There were examples, where were shown how with mathematical induction prove equations like $(1+q)(1+q^2)(1+q^4)\dots(1+q^{{2}^{n}}) = \frac{1-q^{{2}^{n+1}}}{1-q}$.
Now I'm trying to complete exercise where I have to find sum of $\frac{1}{1+x^2}+\frac{1}{(1+x^2)^2}+...+\frac{1}{(1+x^2)^n}$.
I tried to do it with mathematical induction. Like this:
n=1: $\frac{1}{1+x^2}$
n=2: $\frac{1}{1+x^2}+\frac{1}{(1+x^2)^2}=\frac{1+1+x^2}{(1+x^2)^2}$
n=3: $\frac{1}{1+x^2}+\frac{1}{(1+x^2)^2}+\frac{1}{(1+x^2)^3}=\frac{(1+x^2)^2+2+x^2}{(1+x^2)^3}$
...
And so on (I've calculated til n=5). But I don't see any consistent pattern to evaluate sum of progression.
After that I found formulas of geometrical progression:
$q=\frac{b_{n+1}}{b_n}$ and $S_n=\frac{b_1(1-q^n)}{1-q}$, so I've evaluated:
$q=\frac{\frac{1}{(1+x^2)^2}}{\frac{1}{1+x^2}}=\frac{1}{1+x^2}$
and
$S_n=\left(\frac{1}{1+x^2}\left[1-\left(\frac{1}{1+X^2}\right)^n\right]\right):\left(1-\left[\frac{1}{1+x^2}\right]\right)=\left[\frac{1}{1+x^2}-\left(\frac{1}{1+x^2}\right)^{n+1}\right]\frac{1+x^2}{x^2}=\frac{\left[1-\left(\frac{\sqrt[n+1]{1+x^2}}{1+x^2}\right)^{n+1}\right]}{x^2}$
First of all, I'd like to know how to find sum of geometrical progression with mathematical induction.
Secondly, I'd like to know what is wrong with my evaluations.

You like to prove $$q+q^2+\dots+q^n=\frac{q(1-q^n)}{1-q}.$$ using Mathematical induction.
The expression is true for $n=1$ because it is simply $q=q$
If true for n, you want to show that it is also true for $n+1$
Note that $$  q+q^2+\dots+q^n+q^{n+1} = \frac{q(1-q^n)}{1-q} + q^{n+1} = $$
$$ \frac {q(1-q^n)+(1-q)q^{n+1}}{1-q}=  \frac {q(1-q^{n+1})}{1-q} $$ 
With $q= \frac {1}{1+x^2}$ the expression under question simplifies to  $$\frac{1}{1+x^2}+\frac{1}{(1+x^2)^2}+…+\frac{1}{(1+x^2)^n}=\frac {(1+x^2)^n-1}{x^2(1+x^2)^n}$$ 
