For every integer m from 1 to 100, inclusive, the mth term

1st term \((-1)^1*2^{-1}=-\frac{1}{2}\);
2nd term \((-1)^2*2^{-2}=\frac{1}{4}\);
3rd term \((-1)^3*2^{-3}=-\frac{1}{8}\);
4th term \((-1)^4*2^{-4}=\frac{1}{16}\);
5th term \((-1)^5*2^{-5}=-\frac{1}{32}\);
6th term \((-1)^6*2^{-6}=\frac{1}{64}\);
...

The sum of 1st and 2nd = \(-\frac{1}{4}\);
The sum of 3rd and 4th = \(-\frac{1}{16}\);
The sum of 5th and 6th = \(-\frac{1}{64}\);
...

So, we should basically sum 100/2=50 terms which form a geometric progression with the first term of \(-\frac{1}{4}\) and common ratio of \(\frac{1}{4}\): \(-\frac{1}{4}\), \(-\frac{1}{16}\), \(-\frac{1}{64}\), ...

The sum of infinite geometric progression with common ratio \(|r|<1\), is \(sum=\frac{b}{1-r}\), where \(b\) is the first term.

So, in our case, since the # of terms is large enough, the sum will be very close to \(sum=\frac{-\frac{1}{4}}{1-\frac{1}{4}}=-\frac{1}{3}\).

Answer: C.