Impossible problem

Algebra Level 3

Let $f\left( x \right)$ defined on R, $f\left( x \right) =\frac { { 4 }^{ x } }{ { 4 }^{ x }+2 }$ . What is the simplified value of $f\left( \frac { 1 }{ n } \right) +f\left( \frac { 2 }{ n } \right) +...+f\left( \frac { n-1 }{ n } \right) ?$

This problem is part of the set Hard Equations

\frac { 1 }{ 2 } (n+1)

\frac { n }{ 2 }

\frac { 1 }{ 2 } (n-1)

\frac { 4 }{ n+1 }

1 solution

Julian Uy
Dec 5, 2014

$f(x)+f(1-x)=\frac { { 4 }^{ x } }{ 4^{ x }+2 } +\frac { { 4 }^{ 1-x } }{ { 4 }^{ 1-x }+2 }$

$=\frac { { 4 }^{ x } }{ 4^{ x }+2 } +\frac { { 4 } }{ { 4 }+2*{ 4 }^{ x } }$

Therefore,

$f(\frac { 1 }{ n } )+f(\frac { 2 }{ n } )+...+f(\frac { n-1 }{ n } )=\frac { 1 }{ 2 } (n-1)$

since you add the first and last term, the second and second last term, etc.

Impossible problem

1 solution

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