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f(x) = 9^x/ ( 1 + 9^x) then value of f(1...

`f(x) = 9^x/ ( 1 + 9^x)` then value of `f(1/2015) + f(2/2015) .......+ f(4029/2015)`

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`f(x) = 9^x/(9+9^x) = 1/(1+9^(1-x))`
`:. f(1/2015) = 1/(1+9^(2014/2015))`
Now, `f(4029/2015) = 1/(1+9^(-2014/2015)) = (9^(2014/2015))/(1+9^(2014/2015))`
`:. f(1/2015) + f(4029/2015) = 1/(1+9^(2014/2015)) + (9^(2014/2015))/(1+9^(2014/2015)) = 1`
So, first and last term sum is `1`. Similarly, sum of second and second last terms will also be `1` and so on.
As, there are odd number of terms, middle term will be,
`f(2015/2015) = 1/2`
So, the required sum ` = 4028/2+1/2 = 2014.5`
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