Simple question

Using binomial expansion for $(1+x)^n$ , we have

$\displaystyle (1+x)^n = \sum_{k=0}^n \dbinom{n}{k} x^k$

Note that

$\dbinom{n}{k} = \dbinom{n}{n-k}$

Thus, for odd $n$ we can write

$\displaystyle (1+x)^n = \sum_{k=0}^n \dbinom{n}{k} x^k = 2 \sum_{k=0}^{ {(n-1)}/{2} } \dbinom{n}{k} x^k$

Putting $x=1$ and $n=2017$ gives

$\begin{aligned} & \displaystyle 2^{2017} = 2 \sum_{k=0}^{1008} \dbinom{2017}{k} \\ \implies & \displaystyle 2^{2017} = 2 \left( \sum_{k=1}^{1008} \dbinom{2017}{k} + 1 \right) \\ \implies & \displaystyle \sum_{k=1}^{1008} \dbinom{2017}{k} = \dfrac{2^{2017}-2}{2} = \boxed{2^{2016}-1} \end{aligned}$