Finding the sum $\sum_{q=0}^{2016}\sqrt{2}^{\binom{q+3}{3}}$ 
Find the sum
  $\displaystyle{\quad%
\sum_{q = 0}^{2016}\left(\,\sqrt{\, 2\, }\,\right)^{\binom{\displaystyle{q + 3}}{\displaystyle{3}}}}
$.

This sum was encountered while solving a problem of combinatorics. I"ve never seen anything like this before so I'd be grateful if any hints or partial solutions to this problem are given.
Note: I am trying to find the number of  polynomials with degree $q\leq 2016$ and coefficients $1$ or $0$ such that the sum of coefficients is always an odd number. Does the above-written sum solve the problem ?.
 A: No, it doesn't solve that problem.
That problem (not the sum, the one about polynomials) is equivalent to counting  ${\bf x} \in \{0,1\}^{2017}$ such that $\sum_i x_i$ is odd.  Hint: what happens when you flip $x_1$?
A: No, the sum you've written is not helpful for the problem.
First of all, note that there is this useful identity for all $n\in\Bbb{N}$ that $$\sum_{k=0\\\mathrm{even}}^n\binom nk=\sum_{k=1\\\mathrm{odd}}^n\binom nk=2^{n-1}$$
which you can easily prove by expanding $(1+x)^n$ for both $x=-1$ and $x=1$ and grouping even/odd parts.
Now, assume $q$ is given. We're looking for polynomials of degree $q$ with coefficients being $1$ or $0$, and their sum being an odd number. Obviously, the coefficient of the largest degree ($x^q$) has to be $1$ which means the sum of the remaining $q$ coefficients must be even. In other words, we have to choose an even number of coefficients and set them to $1$, and others to $0$. The number of ways we can do this is equal to the number of ways we can choose an even number of the remaining $q$ coefficients, i.e., $$n_q=\sum_{k=0\\\mathrm{even}}^q\binom qk=2^{q-1}$$
Since you want all degrees $q\leq2016$, you have to add up these as (note that I've excluded the case $q=0$ from the summation and added it manually) $$n=1+\sum_{q=1}^{2016}2^{q-1}$$ which equals \begin{aligned}n&=1+2^{2016}-1\\&=2^{2016}\end{aligned}
You can get this answer in a shorter way by using the Robert Israel's answer.
