# NMTC 2000

I was doing this problem

Let $\large\ f(x)=\frac { { 16 }^{ x } }{ { 16 }^{ x }+4 }$

Evalute the sum :

$\large\ f\left( \frac { 1 }{ 2000 } \right) +f\left( \frac { 2 }{ 2000 } \right) +...+f\left( \frac { 1999 }{ 2000 } \right)$

I replaced $x$ by $1-x$ and got

$\large\ f(1-x)=\frac { { 16 }^{ 1-x } }{ { 16 }^{ 1-x }+4 } =\frac { { 16 }/{ 16 }^{ x } }{ (16+4.{ 16 }^{ x })/{ 16 }^{ x } } =\frac { 16 }{ 16+4.{ 16 }^{ x } } =\frac { 4 }{ { 16 }^{ x }+4 }$

I observed that

$\large\ f(x)+f(1-x)=\frac { { 16 }^{ x } }{ { 16 }^{ x }+4 } +\frac { 4 }{ { 16 }^{ x }+4 } =1$.

Thus, i wrote the given summation of functions as :

$\large\ \left[ f\left( \frac { 1 }{ 2000 } \right) +f\left( \frac { 1999 }{ 2000 } \right) \right] +\left[ f\left( \frac { 2 }{ 2000 } \right) +f\left( \frac { 1998 }{ 2000 } \right) \right] +...+\left[ f\left( \frac { 999 }{ 2000 } \right) +f\left( \frac { 1001 }{ 2000 } \right) \right] +f\left( \frac { 1000 }{ 2000 } \right)$

$\large\ =\quad 999\quad \times \quad 1\quad +\quad f\left( \frac { 1 }{ 2 } \right)$

$\large\ =\quad 999\quad +\quad \frac { 1 }{ 2 }$

$\large\ =\quad 999\frac { 1 }{ 2 }$.

Where is wrong?

If you have more logical , easier solution, do post that. Note by Priyanshu Mishra
5 years, 10 months ago

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- 5 years, 9 months ago

its the easiest way and i am pretty sure that its correct

- 5 years, 9 months ago

Yes it is...correct!!

- 5 years, 9 months ago