What is the quotient when $10^n +1$ is divided by $11$, $n$ odd? We have to find the quotient when $10^n + 1$ is divided by $11$, $n$ is odd. 
Now writing $10^n + 1$ as $(11-1)^n + 1$ and expanding , i get the quotient as$Q=\sum_{r=1}^{n}{(-1)}^{r+1}\binom{n}{r}\cdot{11}^{r-1}$ but i have to prove that this $Q$ is actually equal to $9090...9091$ containing $n-1$ digits.
How do I prove it?

$$\begin{align}\frac{1+ x^{2n+1}}{1+x} &= 1 -x + x^2+\cdots + x^{2n}\\[-1em]
&= 1 + (x^2\!-\!x) + (x^4\!-\!x^3) + \cdots + x^{2n}\!-x^{2n-1}\\[.1em]
&= 1 + (x^2\!-\!x)\,(1 + x^2 + x^4 + \cdots + x^{2n-2})\\[.1em]
&= 1 + 90(1010\cdots 0101)\ \ \ {\rm for}\ \ \ x=10 \\[.1em]
&=\qquad\! 9090\cdots 91
\end{align}$$
