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If $$f(x) = \frac{4^x}{4^x+2} $$ then find the value of $$f\left(\frac{1}{1999}\right) + f\left(\frac{2}{1999}\right) + f\left(\frac{3}{1999}\right) +\cdots+f\left(\frac{1999}{1999}\right).$$

I tried it by changing expression to $$f(x) =1 - \frac{2}{4^x+2}$$ but I am not able to cancel any term.

Is there is any other trick to do it? Thanks in advance.

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Hm.. If $z = 4^{1/1999},$ then the sum can be rewritten as $\displaystyle\sum_{i=1}^{1999} \frac{z^i}{z^i+2}.$ –  user2468 Mar 22 '12 at 17:00
    
@vikii did you find this question in Arihant Differential calculus? –  Tomarinator Mar 22 '12 at 17:16
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answer is $999+\frac{2}{3}$ –  Jeremy Carlos Mar 22 '12 at 17:25
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This is not a popularity contest. Relax –  Jeremy Carlos Mar 22 '12 at 19:16
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@JeremyCarlos I intended that as a compliment to you. So, your comment comes to me as rather rude. –  user21436 Mar 22 '12 at 19:47
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1 Answer

up vote 13 down vote accepted

Hint:

Observe

$$f(x)+f(1-x)=1$$

So, $$\begin{align}&f\left(\dfrac 1 {1999}\right)+f\left(\dfrac 2 {1999}\right)+ \cdots+f\left(\dfrac{1998}{1999}\right)+f\left(\dfrac{1999}{1999}\right)\\&=f\left(\dfrac 1 {1999}\right)+f\left(\dfrac{1998}{1999}\right)+f\left(\dfrac 2 {1999}\right)+f\left(\dfrac {1997} {1999}\right)+\cdots +f\left(\dfrac{999}{1999}\right)+f\left(\dfrac{1000}{1999}\right)+f(1)\\&=999+f(1)\\&=999+\dfrac 2 3\\&=\dfrac {2999} 3\end{align}$$


$$\begin{align}f(x)+f(1-x)&=\dfrac{4^x}{4^x+2}+\dfrac{4^{1-x}}{4^{1-x}+2}\\&=\dfrac{4^x}{4^x+2}+\dfrac{\dfrac{4}{4^x}}{\dfrac{4}{4^x}+2}\\&=\dfrac{4^x}{4^x+2}+\dfrac 4 {2 \cdot4^x+4}\\&=\dfrac{4^x}{4^x+2}+\dfrac{\not{4}~~2}{\not2(4^x+2)}\\&=\dfrac{4^x+2}{4^x+2}\\&=1\end{align}$$

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Thanks..:) Got it. –  vikiiii Mar 22 '12 at 17:15
    
i have already mark your answer as accepted.why you worried? –  vikiiii Mar 22 '12 at 17:31
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This is a strange dialogue. –  The Chaz 2.0 Mar 22 '12 at 17:36
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And as rather rude –  Kirthi Raman Mar 22 '12 at 19:54
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@kannappan no.but your comment to jeremy carlos made me think so –  vikiiii Mar 23 '12 at 2:46
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