Can you attack this function?

Algebra Level 4

If $f(x)=\frac{4^x}{4^x+2}$ , where $x€Q$ , then $f(\frac{1}{2007})+f(\frac{2}{2007})+...+f(\frac{2006}{2007})$ equals to $?$

1999 2016 none of them 1003 2015 2007

1 solution

Gabriel Ferreira
Nov 22, 2015

its two diferent ways, integrate or just with basic math. at frist you cam do: 1+2+3+...+2006 its equal to 2006/2(2006+1) so, as de comom divisor is 2007, its just it:

1003x2007/2007 = 1003

Find f(x)+f(1-x). You get that sum to be equal to 1. Therefore f(1/2007)+f(2006/2007) = 1 and similiarly it goes on, to get a solution upto 2006/3. Hence the answer is 1003.

Pavan Gurudath - 5 years, 6 months ago

thanks for the explanation, Pavan!

Florentina Dumitru - 5 years, 6 months ago

but that is the sum of the values of x, not the sum of the functions...

Florentina Dumitru - 5 years, 6 months ago

its the sum of F(x), not X

Gabriel Ferreira - 5 years, 6 months ago

Can you attack this function?

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