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[Solved] Finding the solution of exponential funtion

 
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June 15, 2021 11:42 am
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Staff 
(@armor)
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Let f(x)=\frac{4^x}{4^x+2},   compute   f(\frac{1}{1001})+f(\frac{2}{1001})+...+f(\frac{1000}{1001}).

 

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June 15, 2021 11:42 am
Topic starter
Staff 
(@armor)
Admin
  Honorable Member
317 Posts
156 153 0

Since  \small f(1-x)=\frac{4^{1-x}}{4^{1-x}+2}=\frac{4}{4+2*4^x}=\frac{2}{4^x+2},  then  

\small f(x)+f(1-x)=\frac{4^x}{4^x+2}+\frac{2}{4^x+2}=1

Thus 

\small f(\frac{1}{1001})+f(\frac{2}{1001})+...+f(\frac{1000}{1001})=f(\frac{1}{1001})+f(\frac{1000}{1001})+f(\frac{2}{1001})+f(\frac{999}{1001})+...+f(\frac{500}{1001})+f(\frac{501}{1001})

\small =1+1+1+....+1 (500 times)=500.

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