A nice concept!

Algebra Level 4

If f ( x ) = 4 x 4 x + 2 f(x)= \dfrac{4^{x}}{4^{x}+2} , where x x is a rational number, what is f ( 1 2007 ) + f ( 2 2007 ) + f ( 3 2007 ) + + f ( 2006 2007 ) f \left(\dfrac{1}{2007}\right)+f \left(\dfrac{2}{2007}\right)+f \left(\dfrac{3}{2007}\right)+ \cdots +f \left(\dfrac{2006}{2007}\right) ?


The answer is 1003.

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2 solutions

Chew-Seong Cheong
Oct 22, 2019

Consider the following:

f ( x ) + f ( 1 x ) = 4 x 4 x + 2 + 4 1 x 4 1 x + 2 Multiply up and down by 4 x 1 2 = 4 x 4 x + 2 + 4 1 2 4 1 2 + 2 × 4 x 1 2 = 4 x 4 x + 2 + 2 2 + 4 x = 1 \begin{aligned} f(x) + f(1-x) & = \frac {4^x}{4^x+2} + \blue{\frac {4^{1-x}}{4^{1-x}+2}} & \small \blue{\text{Multiply up and down by } 4^{x-\frac 12}} \\ & = \frac {4^x}{4^x+2} + \blue{\frac {4^\frac 12}{4^\frac 12+2\times 4^{x-\frac 12}}} \\ & = \frac {4^x}{4^x+2} + \frac 2{2+4^x} \\ & = 1 \end{aligned}

Now, we have:

S = f ( 1 2007 ) + f ( 2 2007 ) + f ( 3 2007 ) + + f ( 2004 2007 ) + f ( 2005 2007 ) + f ( 2006 2007 ) = f ( 1 2007 ) + f ( 2 2007 ) + f ( 3 2007 ) + + f ( 1 3 2007 ) + f ( 1 2 2007 ) + f ( 1 1 2007 ) = n = 1 1003 [ f ( n 2007 ) + f ( 1 n 2007 ) ] = n = 1 1003 1 = 1003 \begin{aligned} S & = f \left(\frac 1{2007} \right) + f \left(\frac 2{2007} \right) + f \left(\frac 3{2007} \right) + \cdots + f \left(\frac {2004}{2007} \right) + f \left(\frac {2005}{2007} \right) + f \left(\frac {2006}{2007} \right) \\ & = f \left(\frac 1{2007} \right) + f \left(\frac 2{2007} \right) + f \left(\frac 3{2007} \right) + \cdots + f \left(1-\frac 3{2007} \right) + f \left(1-\frac 2{2007} \right) + f \left(1 - \frac 1{2007} \right) \\ & = \sum_{n=1}^{1003} \left[ f \left(\frac n{2007} \right) + f \left(1-\frac n{2007} \right)\right] = \sum_{n=1}^{1003} 1 = \boxed{1003} \end{aligned}

Surya Prakash
Aug 1, 2015

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

So,

f ( 1 2007 ) + f ( 2 2007 ) + . . . . . . . + f ( 2006 2007 ) f(\dfrac{1}{2007}) +f(\dfrac{2}{2007}) +.......+f(\dfrac{2006}{2007}) = ( f ( 1 2007 ) + f ( 2006 2007 ) ) + . . . . . . . . + ( f ( 1003 2007 ) + f ( 1004 2007 ) ) =(f(\dfrac{1}{2007})+f(\dfrac{2006}{2007}))+........+(f(\dfrac{1003}{2007})+ f(\dfrac{1004}{2007})) = 1 + 1 + 1 + . . . . . . . ’1003’ times . . . . . + 1 = 1003 =1+1+1+.......\text{'1003' times} .....+1=\boxed{1003}

Moderator note:

Simple standard approach.

How can one guess that f ( x ) + f ( 1 x ) = 1 f(x) + f(1-x) = 1 ?

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