Cool functions

$Note\quad that\quad f(1-x)+f(x)=\frac { { a }^{ 1-x } }{ { a }^{ 1-x }+\sqrt { a } } +\frac { { a }^{ x } }{ { a }^{ x }+\sqrt { a } } =\frac { a+{ a }^{ \frac { 3 }{ 2 } -x }+{ a }^{ x }+{ a }^{ x+\frac { 1 }{ 2 } } }{ a+{ a }^{ \frac { 3 }{ 2 } -x }+{ a }^{ x }+a^{ x+\frac { 1 }{ 2 } } } =1\\ \\ Now\quad \sum _{ r=0 }^{ 4n }{ f\left( \frac { r }{ 4n+1 } \right) } =\frac { 1 }{ 1+\sqrt { a } } +1974\quad \equiv \sum _{ r=1 }^{ 4n }{ f\left( \frac { r }{ 4n+1 } \right) } =1974\quad \left\{ \because f(0)=\frac { 1 }{ 1+\sqrt { a } } \right\} \\ \\ \sum _{ r=1 }^{ 4n }{ f\left( \frac { r }{ 4n+1 } \right) } =\left\{ \left[ f\left( \frac { 1 }{ 4n+1 } \right) +f\left( \frac { 4n }{ 4n+1 } \right) \right] +\left[ f\left( \frac { 2 }{ 4n+1 } \right) +f\left( \frac { 4n-1 }{ 4n+1 } \right) \right] ...............2n\quad times \right\} =2n\\ \therefore 2n=1974\Rightarrow \boxed { n=987 } \\$