Question 1197013
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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
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<pre>
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^(1-x)/(3+9^(1-x))}}} = 

           = {{{9/(9^x*(3+9/9^x)))}}} = {{{(9*9^x)/(9^x*(3*9^x+9))}}} = {{{9/(3*9^x+9)}}} = {{{3/(9^x+3)}}} = {{{3/(3+9^x)}}}.


    THEREFORE,  f(x) + f(1-x) = {{{9^x/(3 + 9^x)}}} + {{{3/(3+9^x)}}} = {{{(9^x+3)/(3+9^x)}}} = 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(i/2016)}}} + {{{f(1-i/2016)}}}, 
and the sum in each such a pair equals 1.


So, the doubled sum equals 2015 and the sum itself is  {{{2015/2}}},  exactly as the problem states.
</pre>

Q.E.D.    &nbsp;&nbsp;&nbsp;&nbsp;Solved.