A binary exansion

Express the number 101_{\dot 2}10\overline{011} as a fraction.

Looking at the values left of the decimal point, we have:
(1)(2^0)+(0)(2^1)+(1)(2^2)=5

Looking at the left side of the decimal point, we have:
(1)(2^{-1})+(0)(2^{-2})+(0)(2^{-3})+(1)(2^{-4})+(1)(2^{-5})+(0)(2^{-6})+(1)(2^{-7})+.....

The fact that it is a repeating decimal point complicates things, but there is still hope.
First, let us look rewrite the sum like so:
\frac{1}{2}+\frac{1}{2^4}+\frac{1}{2^5}+\frac{1}{2^7}+\frac{1}{2^8}+\frac{1}{2^{10}}+....
Factoring out from the sum we obtain:
\frac{1}{2}+\frac{1}{2^4} \bigg(1+\frac{1}{2}+\frac{1}{2^3}+\frac{1}{2^4}+.......\bigg)
Which is:
\frac{1}{2}+\frac{1}{2^4} \bigg(\sum_{n=0}^{\infty} \frac{1}{2^n}-\sum_{n=0}^{\infty}\frac{1}{2^{3n+2}} \bigg)

Thus we can rewrite the whole thing as:
5+\frac{1}{2}+\frac{1}{2^4} \bigg(\sum_{n=0}^{\infty} \frac{1}{2^n}-\sum_{n=0}^{\infty}\frac{1}{2^{3n+2}} \bigg)

Using the Geometric Series to calculate the summation terms:
5+\frac{1}{2}+\frac{1}{2^4} \big(\frac{12}{7}\big)
Writing this as a fraction:
\frac{(5)(2^4)(7)+(2^2)(7)+12}{(2^4)(7)}=\frac{23}{4}

Way to take all the good ones cornelius

I made a mistake. The correct answer is \frac{157}{28}.

The 5+\frac{1}{2} part is correct, however the rest is not.

Let us try to convert 0.000\overline{011} to a fraction.
First, let us write out some values:

\frac{1}{2^4}+\frac{1}{2^5}+\frac{1}{2^7}+\frac{1}{2^8}+\frac{1}{2^{10}}+...

Let’s factor out a \frac{1}{2^4}:
\frac{1}{2^4} \bigg(1+\frac{1}{2}+\frac{1}{2^3}+\frac{1}{2^4}+\frac{1}{2^6}+...\bigg)

To simplify, we will add the consecutive couples together:
\frac{1}{2^4} \bigg(1+\frac{1}{2}+\frac{1}{2^3}+\frac{1}{2^4}+\frac{1}{2^6}+...\bigg)=\frac{1}{2^4}\bigg(\frac{3}{2}+\frac{3}{2^4}+\frac{3}{2^7}+...\bigg)
=\bigg(\frac{1}{2^4}\bigg) \bigg(\frac{3}{2}\bigg) \bigg(1+\frac{1}{2^3}+\frac{1}{2^6}+...\bigg)
=\bigg(\frac{1}{2^4}\bigg) \bigg(\frac{3}{2}\bigg) \bigg(\sum_{n=0}^{\infty}\bigg(\frac{1}{2^3}\bigg)\bigg)=\frac{3}{2^5}\bigg(\frac{8}{7}\bigg)=\frac{3}{28}

Putting it all together:
5+\frac{1}{2}+\frac{3}{28}=\frac{157}{28}