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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?

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4 Answers 4

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

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  • $\begingroup$ Oh, then I should have paired up with $f(-5)$ and $f(6)$ instead. $\endgroup$
    – Anonymous
    Jan 2, 2021 at 18:09
  • $\begingroup$ @Anonymous Yes, exactly! $\endgroup$
    – Robert Z
    Jan 2, 2021 at 18:14
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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: $$\frac{1}{3^x+\sqrt 3}+\frac{1}{3^{1-x}+\sqrt 3}=\frac{3^x+3^{1-x}+2\cdot 3^\frac 12}{3^{x+\frac 12}+3^{\frac32-x}+2\cdot 3^1}=\frac{\alpha}{3^\frac 12 \cdot \alpha}=\frac{1}{\sqrt 3}$$

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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?

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