Let `f(x)=(9^x)/(9^x+3)` . Show `f(x)+f(1-x)=1` and, hence, evaluate. `f(1/(1996))+f(2/(1996))+f(3/(1996))++f((1995)/(1996))`

Explanation

`f(x)=(9^(x))/(9^(x)+3) " (1)" `
and `f(1-x)=(9^(1-x))/(9^(1-x)+3).`
or `f(1-x)=((9)/(9^(x)))/((9)/(9^(x))+3)=(9)/(9+3.9^(x))`
or `f(1-x)=(3)/((3+9^(x))) " (2)" `
Adding (1) and (2), we get
`f(x)+f(1-x)=(9^(x))/(9^(x)+3)+(3)/((3+9^(x)))=1`
or `f(x)+f(1-x)=1 " (3)" `
Now putting `x=(1)/(1996),(2)/(1996),(3)/(1996), ..., (998)/(1996) ` in (3) we get
`f((1)/(1996))+f((1995)/(1996))=1, f((2)/(1996))+f((1994)/(1996))=1,`
`f((3)/(1996))+f((1993)/(1996))=1`
` " " vdots`
`f((997)/(1996))+f((999)/(1996))=1, f((998)/(1996))+f((998)/(1996))=1`
` :. f((998)/(1996))=(1)/(2)`
Adding all the above expressions, we get
`f((1)/(1996))+f((2)/(1996))+ ... +f((1995)/(1996))`
` = (1+1+1+ ... +997" times")+(1)/(2)`
`=997+(1)/(2)=997.5`