The important redundancy

We have $f\left( x \right) =\frac { { 9 }^{ x } }{ { 9 }^{ x }+3 } ,f(1-x)=\frac { { 9 }^{ 1-x } }{ { 9 }^{ 1-x }+3 } =\frac { 9 }{ 9+3\times { 9 }^{ x } }$ .

$f\left( x \right) +f(x-1)=\frac { 9 }{ 9+3\times { 9 }^{ x } } +\frac { { 9 }^{ x } }{ { 9 }^{ x }+3 } =1$ by cross multiplying a simplifying.

$\therefore f\left( \frac { n }{ 1996 } \right) +f\left( \frac { 1996-n }{ 1996 } \right) =1$ .

Mean of $1,2,3,....,1995$ is $998$ .

$b=f\left( \frac { 1 }{ 1996 } \right) +f\left( \frac { 2 }{ 1996 } \right) +...+f\left( \frac { 1995 }{ 1996 } \right)$

$=1+1+1+..+1$ (997 times) + $f\left( \frac { 998 }{ 1996 } \right)$ .

We know that $\therefore f\left( \frac { 998 }{ 1996 } \right) +f\left( \frac { 1996-998 }{ 1996 } \right) =1=2f\left( \frac { 998 }{ 1996 } \right)$ .

$\therefore f\left( \frac { 998 }{ 1996 } \right) =0.5$ .

Therefore $b=997.5$ and $a=1$ .

9^x can be assumed as "t"

f (x) = t/(t+3) ; f (1-x) = (9/t)/(9/t + 3) { since 9^(1-x) = 9/t } .

therefore, a = f (x) + f (1-x) = t/(t+3) + 3/(t+3) = 1 .

f (1995/1996) can be written