TITLE: summation of a binomial expression that doesn't start from 0
QUESTION [6 upvotes]: 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

REPLY [10 votes]: 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.