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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2 Answers

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
@Noam: You are welcome! – Remark: You got some answers to your questions now. Please don't forget to accept helpful answers eventually. For each of your questions, try to choose one answer and accept it by clicking on the check mark. That marks the problem as closed and gives some reputation points to you and to the author of the answer. See math.stackexchange.com/help/someone-answers for more information. – Martin R Jan 31 at 15:31

Consider

$$\sum_{k=0}^{17}\binom{17}{k} -\sum_{k=0}^{8}\binom{17}{k}= 2^{17} - 2^{16}$$

$$=2^{16}(2-1)= 2^{16}$$

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