Quiz problem: A binary expansion

Express $10_{\dot 2}1\overline{101}$ as a fraction.

Here goes nothing...

$$ 10.1\overline{101} = 2 + 0 + \frac{1}{2}+ \frac{1}{2}\sum_{n=1}^{\infty} \frac{3}{5^k} $$

$$ = 2 + \frac{1}{2} + \frac{1}{2}(\frac{3}{2}) $$

$$ = 2 + \frac{1}{2} + \frac{3}{4} $$

$$ = \frac{13}{4} $$

The three in the numerator of the geometric sum comes from the number of digits repeating. The five in the denominator comes from the fact that 101 is binary for 5.

@djett Looks good so far. I did fix the TeX a bit. Most importantly, you don't need to indent it four times - otherwise, it displays as code, rather than typesetting.

As far as the math goes, it's not quite finished. You've correctly expressed the number as a series but can you go one more step and apply the geometric series formula?

Here's what I'm getting:

$$10.1\overline{101} = 2+0+ \frac{1}{2} + \frac{1}{2}\sum _{k=1}^\infty \frac{5}{2^{3k}}$$
$$ = 2 + \frac{1}{2} + \left(\frac{1}{2}\right)\left(\frac{(5)(\frac{1}{8})^1}{1-\frac{1}{8}}\right)$$
$$=\frac{5}{2} + \left(\frac{1}{2}\right)\left(\frac{5}{8}\right)\left(\frac{8}{7}\right)$$
$$=\frac{5}{2} + \frac{5}{14} = \frac{20}{7}.$$

The numerator must be $5$, and it repeats every $2^{3k}$ after the initial shift.

So I'm just trying to confirm that my thought process for this kind of problem is correct, but when you have
$$10_{\dot 2}1\overline{101}=2+0+1+\frac{1}{2}+\frac{1}{2}\sum_{k=1}^\infty \frac{5}{2^{3k}},$$
the $\frac{1}{2}$ preceding the summation comes from the fact that we have $k=1$ for our sum and are shifting everything down by a power of $\frac{1}{2}$, correct?

So the following two statements should be equivalent?

$$\frac{1}{2}\sum_{k=1}^\infty \frac{5}{2^{3k}}=\sum_{k=2}^\infty \frac{5}{2^{3k}}$$

@Shia No, not quite. If you plug in $k=1$ on the left and account for the $1/2$ outside the sum, then you can see that your first term is
$$\frac{1}{2} \frac{5}{2^3} = \frac{5}{2^4}.$$
On the right, though, the first term is
$$\frac{5}{2^{3\times2}} = \frac{5}{2^6}.$$
I think the correct way to distribute the $1/2$ is just to add one to the exponent inside. So
$$\frac{1}{2}\sum_{k=1}^\infty \frac{5}{2^{3k}}=\sum_{k=1}^\infty \frac{5}{2^{3k+1}}.$$

Okay, so then does the $\frac{1}{2}$ come from the repeating sequence being shifted one spot further to the right from the one's place?

As in, if our sequence were something like
$$10_{\dot 2}\overline{101},$$

Then we would say that it equates to
$$2+0+1+\sum_{k=1}^\infty \frac{5}{2^{3k}}?$$

@Shia

Yes, I believe the $\frac{1}{2}$ comes from the shift of the repeating part of the function.

I think though in your example Im not sure where the +1 came from I think it should be

$$2+0+\sum_{k=1}^{\infty}\frac{5}{2^{3k}}$$

I could be wrong but I think that is correct.