<s>[INST] Consider the function $f(x) = \frac{1}{3^x + \sqrt{3}}.$ Find :- $\sqrt{3}[f(-5) + f(-4) + ... + f(3) + f(4) + f(5) + f(6)]$. 
Consider the function $f(x) = \frac{1}{3^x + \sqrt{3}}.$ Find :-
$\sqrt{3}[f(-5) + f(-4) + ... + f(3) + f(4) + f(5) + f(6)]$.

What I Tried: I checked similar questions and answers in the Art of Problem Solving here and here and tried to get some ideas.
First thing which I did is thinking of pairing the values, I took for example, $f(-1)$ and $f(1)$.
We have :-
$$\rightarrow f(-1) = \frac{1}{\frac{1}{3} + \sqrt{3}} = \frac{3\sqrt{3} + 1}{3}$$
$$\rightarrow f(1) = \frac{1}{3 + \sqrt{3}}$$
Adding both gives $\frac{7 + 6\sqrt{3}}{12 + 10\sqrt{3}}$, which more or less looks like a random sum.
So my idea of pairing did not work, or at least I couldn't pair them nicely or missed a pattern. So how would I start solving it?
Can anyone help?
 [/INST] $\sqrt{3}f(x) = \frac{1}{\sqrt{3^{2x-1}}+1}$ and $\frac{1}{\sqrt{3^{a}}+1}+\frac{1}{\sqrt{3^{-a}}+1}=1$.
You can pair like $(-11,11),(-9,9),(-7,7)\cdots(-1,1)$ then answer is $6$.
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