How can this sum $\sum_{k=0}^a \binom{n}{k}$ be simplified?

Question

I failed to simplify this expression on my exams, not sure if it was even needed, but i wonder if it is possible to simplify $$\sum_{k=0}^{299} \binom{1000}{k}$$ by using $\binom{n}{k}=\binom{n}{n-k}$ and $\sum_{k=0}^n \binom{n}{k}=2^n$, etc.