SOLUTION: If f(x) = 9^x/(3 + 9^x), prove that: f(1/2016)+f(2/2016)+f(3/2016) +... + f(2015/2016)= 2015/2

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Question 1197013: If f(x) = 9^x/(3 + 9^x), prove that:
f(1/2016)+f(2/2016)+f(3/2016) +... + f(2015/2016)= 2015/2

Answer by ikleyn(52802) About Me  (Show Source):
You can put this solution on YOUR website!
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If f(x) = 9^x/(3 + 9^x), prove that:
f(1/2016)+f(2/2016)+f(3/2016) +... + f(2015/2016)= 2015/2
~~~~~~~~~~~~~~~~~~

As first step, let's prove that f(x) + f(1-x) = 1  for any value of x.
We have 

    f(1-x) = by the definition of function f(x) = 9%5E%281-x%29%2F%283%2B9%5E%281-x%29%29 = 

           = 9%2F%289%5Ex%2A%283%2B9%2F9%5Ex%29%29%29 = %289%2A9%5Ex%29%2F%289%5Ex%2A%283%2A9%5Ex%2B9%29%29 = 9%2F%283%2A9%5Ex%2B9%29 = 3%2F%289%5Ex%2B3%29 = 3%2F%283%2B9%5Ex%29.


    THEREFORE,  f(x) + f(1-x) = 9%5Ex%2F%283+%2B+9%5Ex%29 + 3%2F%283%2B9%5Ex%29 = %289%5Ex%2B3%29%2F%283%2B9%5Ex%29 = 1,

    and the statement is proved.



As the next step, let's write two identical sums in direct and inverse order

    f(1/2016)     + f(2/2016)    + f(3/2016)    + . . . + f(2015/2016)

    f(20125/2016) + f(2014/2016) + f(2013/2016) + . . . + f(1/2016)


and add them. Pairing the addends vertically, we have 2015 pairs of the form  f%28i%2F2016%29 + f%281-i%2F2016%29, 
and the sum in each such a pair equals 1.


So, the doubled sum equals 2015 and the sum itself is  2015%2F2,  exactly as the problem states.

Q.E.D.     Solved.