# summation of a binomial expression that doesn't start from 0

I have the following expression:

$$\sum_{k=9}^{17}\binom{17}{k}$$ and I need to show that it's equal to: $$2^{16}$$ now I know that if 'k' was starting from zero and not from 9 , like this: $$\sum_{k=0}^{17}\binom{17}{k}$$ then there is this identity that says it's equal to: $$2^{17}$$ But Because the summation starts from 9 I don't know what to do.. can you help please? thank you

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For integers $n, k$ with $0 \le k \le n$, the binomial coefficients satisfy the “symmetry” $$\binom{n}{k} = \binom{n}{n-k}$$

It follows that $$\sum_{k=0}^{8}\binom{17}{k} = \sum_{k=9}^{17}\binom{17}{17-k} = \sum_{k=9}^{17}\binom{17}{k}$$ and therefore $$\sum_{k=9}^{17}\binom{17}{k} = \frac 12 \sum_{k=0}^{17}\binom{17}{k} = \frac 12 \cdot (1+1)^{17} = 2^{16} \, .$$

But note that this approach works only in this symmetric case where we sum the first or second half of the binomial coefficients in a row of Pascal's triangle for odd $n$ (or with a small modification for even $n$).

According to Wikipedia, there is no closed formula for the general case $\sum_{k=j}^n \binom nk$ unless one resorts to the Hypergeometric function.

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I don't understand why: $$\sum_{k=9}^{17}\binom{17}{k} = 0.5 * \sum_{k=0}^{17}\binom{17}{k}$$ – Noam Jan 31 at 11:25
@Noam: See expanded answer. – Martin R Jan 31 at 11:28
Ok. Now I understand. Thank you very much @Martin R – Noam Jan 31 at 11:32
$$\sum_{k=0}^{17}\binom{17}{k} -\sum_{k=0}^{8}\binom{17}{k}= 2^{17} - 2^{16}$$
$$=2^{16}(2-1)= 2^{16}$$