I am studying for a math test, and I'm wondering on an easier way to add combination series. For example, $12 \choose 3$ + $12 \choose 4$ + ... + $12 \choose 8$. Is there an easier way than: $2^{12}$ - $12 \choose 2$... and etc.
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The only simplification I can think of is maybe exploiting the fact that $\binom{n}{k}=\binom{n}{n-k}$, but that doesn't really make it that much faster.
Sherlock Holmes
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It makes it perhaps a little faster: $$2^{12}-2(1+12+66)-\binom{12}3=4096-158-220=3938-220=3718;,$$ all done as I was typing it out. But basically I agree. – Brian M. Scott Oct 09 '14 at 05:11
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Finding a closed form for this kind of sum is an inherently hard problem. See https://mathoverflow.net/questions/17202/sum-of-the-first-k-binomial-coefficients-for-fixed-n and Bounds on $\sum_{k=0}^{m} \binom{n}{k}x^k$ and $\sum_{k=0}^{m} \binom{n}{k}x^k(1-x)^{n-k}, m<n$.
However, if you have a computer ( and some math software :-) ), you can take advantage of fast methods to evaluate the ordinary hypergeometric function at particular points since
