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Let \(f(x)=\frac{2^x}{2^x+\sqrt{2}}, \text{then }\sum_{k=1}^{81} f\left(\frac{k}{82}\right)\) is equal to

(1) \(\frac{81}{2}\)

(2) 41

(3) \(41 \sqrt{2}\)

(4) 81

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1 Answer

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by (50.3k points)
edited by

Correct option is (1) \(\frac{81}{2}\)   

\(f(x)=\frac{2^x}{2^x+2^{1 / 2}}=\frac{2^x}{2^x+\sqrt{2}}\)   

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

\(=\frac{\sqrt{2}}{2^x+\sqrt{2}} \)  

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

\( \Rightarrow \sum_{k=1}^{81} f\left(\frac{k}{82}\right)+f\left(\frac{2}{82}\right)+\left(f\left(\frac{3}{82}\right)\right)+\ldots \ldots+f\left(\frac{40}{82}\right)+f\left(\frac{41}{82}\right)+f\left(\frac{42}{82}\right) +\ldots+f\left(\frac{79}{82}\right)+f\left(\frac{80}{82}\right)+f\left(\frac{81}{82}\right) \)   

\( =\left[f\left(\frac{1}{82}\right)+f\left(\frac{81}{82}\right)\right]+\left[f\left(\frac{2}{82}\right)+f\left(\frac{80}{82}\right)\right]+\ldots +\left[f\left(\frac{40}{82}\right)+f\left(\frac{42}{82}\right)+f\left(\frac{41}{82}\right)\right] \)    

equation

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