Compute ${11 \choose 1} + {11 \choose 3} + ... + {11 \choose 11}$?

Since any time you choose an odd number of elements out of a set of eleven, you leave behind an even number of elements, $$ \sum_{k\text{ odd}} \binom{11}{k} = \sum_{k\text{ even}} \binom{11}{k} $$

So the number of ways of choosing any subset of eleven elements is twice the number of ways of choosing an odd number of elements.

But to choose any subset, each element is either in our out of the chosen subset: Than gives $2^{11}$.

So the number of ways to choose an odd number of elements is $$\frac12\cdot 2^{11} = 2^{10} = 1024$$

P.S. This reasoning does not work when the main set has an even number of elements. But the property of the number of ways to choose an odd number of elements being half the total number of subsets still holds. Can you see why?

I don't know if this is the fastest method, but you can use the binomial theorem twice: $$2^{11}=(1+1)^{11}=\sum_{i=0}^{11}\binom{11}{i}$$ and $$0=(1-1)^{11}=\sum_{i=0}^{11}\binom{11}{i}(-1)^i.$$ Upon subtracting these two results and dividing by two, we obtain the result.

Since $\binom{11}{1}=\binom{11}{10},\binom{11}{3}=\binom{11}{8}$ and so on, $$ \binom{11}{1}+\binom{11}{3}+\ldots+\binom{11}{11} = \binom{11}{10}+\binom{11}{8}+\ldots+\binom{11}{0} $$ so both the terms are half the sum $\binom{11}{0}+\binom{11}{1}+\binom{11}{2}+\ldots+\binom{11}{11}=2^{11}$, i.e. $2^{10}=\color{red}{1024}$.