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?

Hint: We have that $$f(x)+f(1-x)=\frac{1}{3^{x} + \sqrt{3}}+\frac{1}{3^{1-x} + \sqrt{3}}=\frac{1}{3^{x} + \sqrt{3}}+\frac{3^{x}/\sqrt{3}}{\sqrt{3} + 3^x}=\frac{1}{\sqrt{3}}$$

• Oh, then I should have paired up with $f(-5)$ and $f(6)$ instead. Jan 2, 2021 at 18:09
• @Anonymous Yes, exactly! Jan 2, 2021 at 18:14

You're ever so slightly off. Notice the median of $$(-5, -4, ..., 5, 6)$$ is $$\frac{-5+6}{2}=\frac 12$$ which hints at trying $$f(\frac12+x)+f(\frac 12-x)\overbrace{=}^{y=x+\frac 12}f(y)+f(1-y)$$ We see that: $$\frac{1}{3^x+\sqrt 3}+\frac{1}{3^{1-x}+\sqrt 3}=\frac{3^x+3^{1-x}+2\cdot 3^\frac 12}{3^{x+\frac 12}+3^{\frac32-x}+2\cdot 3^1}=\frac{\alpha}{3^\frac 12 \cdot \alpha}=\frac{1}{\sqrt 3}$$

Let $$\sqrt3=a$$

$$\dfrac{f(x)}a=\dfrac1{a^{2x-1}+1}$$

If $$\dfrac{a^{2x-1}}{1+a^{2x-1}}=f(px+q)=\dfrac1{1+a^{2(px+q)-1}}$$

$$\implies 1-2x=2(px+q)-1$$

$$\implies p=-1,q=1$$

$$\implies f(x)+f(1-x)=a$$

Can you take it from here?

$$\sqrt{3}f(x) = \frac{1}{\sqrt{3^{2x-1}}+1}$$ 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 $$6$$.