So, this is simply a geometric progression, so rewriting I get $$1-\frac{1}{2}+\frac{1}{2^2}-\frac{1}{2^3}+...+\frac{1}{2^{10}}=1-2^{-1}+2^{-2}-2^{-3}+...+2^{-10}$$

$$=\sum\_{i=0}^{10}(-2)^{-i}=\frac{1-(-2^{10+1})}{1-(-2)}=\frac{1+2^{11}}{3}=\frac{2049}{3}=683.$$

Which is incorrect. I used the formula for closed form of geometric sum: $$\sum\_{i=0}^{n}k^{i}=\frac{1-k^{n+1}}{1-k}.$$

In my case $k=-2$ and $n=10$, plugging this in I get the incorrect answer. Can someone point out my misstake?