Let's look for a pattern

If $n=1$, the sequence is $\frac{1}{2}$. So the SUM = $\frac{1}{2}$

If $n=2$, the sequence is $\frac{1}{2}$, $\frac{1}{4}$. So the SUM = $\frac{3}{4}$

If $n=3$, the sequence is $\frac{1}{2}$, $\frac{1}{4}$, $\frac{1}{8}$. So the SUM = $\frac{7}{8}$

If $n=4$, the sequence is $\frac{1}{2}$, $\frac{1}{4}$, $\frac{1}{8}$, $\frac{1}{16}$. So the SUM = $\frac{15}{16}$

At this point, we may notice that the SUM is always less than 1.
So, we can ELIMINATE C, D and E

Now examine the remaining answer choices (A and B).
This gives us a hint about the correct answer.

Notice that, when we have $n$ terms, the SUM is always $\frac{1}{2^n}$ LESS THAN 1

The last term in the given sequence is $\frac{1}{2^{10}}$

So, the sum will equal $1-\frac{1}{2^{10}}$

Answer: A

Cheers,
Brent