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I've been solving a complex problem, in which at one step, I have to calculate the following $$ \frac {{n \choose 0} + {n\choose 1} + {n\choose 2} + ... + {n\choose k}}{2^n} $$ The values of n and k can be very large $ \le 10^5$. But since the answer will always be less than 1 ($ k < n$), I was wondering if its calculable doing some modulo trick which will leave me an answer with 4-5 decimal precision.

Thanks in advance.

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    This may be useful (there is, among others, a closed form using the hypergeometric series): http://mathoverflow.net/questions/17202/sum-of-the-first-k-binomial-coefficients-for-fixed-n – 2017-02-05
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    You could use the similarity to the binomial distribution and approximate it with a normal distribution. And then use that the integrated normal distribution is related to the error function. – 2017-02-05

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