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Question Number 157611 by naka3546 last updated on 25/Oct/21

Given  g(x) = (1/(1 + 3^((1/2) − x) ))  g((1/(2017))) + g((2/(2017))) + g((3/(2017))) + … + g(((2016)/(2017)))  =  ?

$${Given}\:\:{g}\left({x}\right)\:=\:\frac{\mathrm{1}}{\mathrm{1}\:+\:\mathrm{3}^{\frac{\mathrm{1}}{\mathrm{2}}\:−\:{x}} } \\ $$$${g}\left(\frac{\mathrm{1}}{\mathrm{2017}}\right)\:+\:{g}\left(\frac{\mathrm{2}}{\mathrm{2017}}\right)\:+\:{g}\left(\frac{\mathrm{3}}{\mathrm{2017}}\right)\:+\:\ldots\:+\:{g}\left(\frac{\mathrm{2016}}{\mathrm{2017}}\right)\:\:=\:\:? \\ $$

Answered by Ar Brandon last updated on 25/Oct/21

g(x)=(1/(1+3^((1/2)−x) ))=(3^x /(3^x +3^(1/2) ))  g(1−x)=(3^(1−x) /(3^(1−x) +3^(1/2) ))=(3/(3+3^(x+(1/2)) ))=(3^(1/2) /(3^(1/2) +3^x ))  g(x)+g(1−x)=((3^x +3^(1/2) )/(3^x +3^(1/2) ))=1  Σ=((2016)/2)=1008

$$\mathrm{g}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{1}+\mathrm{3}^{\frac{\mathrm{1}}{\mathrm{2}}−{x}} }=\frac{\mathrm{3}^{{x}} }{\mathrm{3}^{{x}} +\mathrm{3}^{\frac{\mathrm{1}}{\mathrm{2}}} } \\ $$$$\mathrm{g}\left(\mathrm{1}−{x}\right)=\frac{\mathrm{3}^{\mathrm{1}−{x}} }{\mathrm{3}^{\mathrm{1}−{x}} +\mathrm{3}^{\frac{\mathrm{1}}{\mathrm{2}}} }=\frac{\mathrm{3}}{\mathrm{3}+\mathrm{3}^{{x}+\frac{\mathrm{1}}{\mathrm{2}}} }=\frac{\mathrm{3}^{\frac{\mathrm{1}}{\mathrm{2}}} }{\mathrm{3}^{\frac{\mathrm{1}}{\mathrm{2}}} +\mathrm{3}^{{x}} } \\ $$$$\mathrm{g}\left({x}\right)+\mathrm{g}\left(\mathrm{1}−{x}\right)=\frac{\mathrm{3}^{{x}} +\mathrm{3}^{\frac{\mathrm{1}}{\mathrm{2}}} }{\mathrm{3}^{{x}} +\mathrm{3}^{\frac{\mathrm{1}}{\mathrm{2}}} }=\mathrm{1} \\ $$$$\Sigma=\frac{\mathrm{2016}}{\mathrm{2}}=\mathrm{1008} \\ $$

Commented by naka3546 last updated on 25/Oct/21

Thank  you,  sir.

$${Thank}\:\:{you},\:\:{sir}. \\ $$

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