How can I calculate $ \sum_{k=35}^{80} {80 \choose k}$?

Question

How can I calculate $$ \sum_{k=35}^{80} {80 \choose k}$$ ?

I don't have idea how to calculate it, and I will be happy to listen some ideas , hints.

Answer 1

You can use Binomial Theorem and observe and the sum up to 40 is about half of a total. So it is enough to have the sum from 35 to 40 and our task seems to be easy. The total from 0 to 80 is of course $2^{80}$.

Answer 2

Not really sure what kind of an expression you are after, but you could notice that $$ \sum_{k=0}^{80} \binom{80}{k} = \sum_{k=0}^{80} \binom{80}{k}1^{k}1^{80-k} = (1+1)^{80} = 2^{80}, $$ and by symmetry, $$ \sum_{k=0}^{39} \binom{80}{k} = \sum_{k=41}^{80} \binom{80}{k}, $$ so that $$ \sum_{k=0}^{80} \binom{80}{k} = 2\sum_{k=41}^{80} \binom{80}{k}+\binom{80}{40}, $$ giving $$ \sum_{k=41}^{80} \binom{80}{k} = \frac{2^{80}-\binom{80}{40}}{2}. $$ Then \begin{align} \sum_{k=35}^{80} \binom{80}{k} &= \binom{80}{35} + \binom{80}{36} + \binom{80}{37} + \binom{80}{38} + \binom{80}{39} + \binom{80}{40} + \sum_{k=41}^{80} \binom{80}{k}\\[0.2cm] &= \binom{80}{35} + \binom{80}{36} + \binom{80}{37} + \binom{80}{38} + \binom{80}{39} + \binom{80}{40} + \frac{2^{80}-\binom{80}{40}}{2}\\[0.2cm] &= \binom{80}{35} + \binom{80}{36} + \binom{80}{37} + \binom{80}{38} + \binom{80}{39} + \frac{1}{2}\binom{80}{40} + 2^{79}. \end{align}

Answer 3

You can use the Gaussian Hypergeometric function:

$$\sum_{k=35}^{80}\dbinom{80}{k} = \dbinom{80}{35}{_2F_1}(1,-45;36;-1)$$

More generally:

$$\sum_{k=r}^n \dbinom{n}{k} = \dbinom{n}{r}{_2F_1}(1,r-n;r+1;-1)$$