×

# 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
1 year, 6 months ago

Sort by:

its the easiest way and i am pretty sure that its correct · 1 year, 5 months ago

Yes it is...correct!! · 1 year, 5 months ago