NMTC 2000

I was doing this problem

Let  f(x)=16x16x+4\large\ f(x)=\frac { { 16 }^{ x } }{ { 16 }^{ x }+4 }

Evalute the sum :

 f(12000)+f(22000)+...+f(19992000)\large\ f\left( \frac { 1 }{ 2000 } \right) +f\left( \frac { 2 }{ 2000 } \right) +...+f\left( \frac { 1999 }{ 2000 } \right)

I replaced xx by 1x1-x and got

 f(1x)=161x161x+4=16/16x(16+4.16x)/16x=1616+4.16x=416x+4\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

 f(x)+f(1x)=16x16x+4+416x+4=1\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 :

 [f(12000)+f(19992000)]+[f(22000)+f(19982000)]+...+[f(9992000)+f(10012000)]+f(10002000)\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)

 =999×1+f(12)\large\ =\quad 999\quad \times \quad 1\quad +\quad f\left( \frac { 1 }{ 2 } \right)

 =999+12\large\ =\quad 999\quad +\quad \frac { 1 }{ 2 }

 =99912\large\ =\quad 999\frac { 1 }{ 2 }.

Where is wrong?

If you have more logical , easier solution, do post that.

Note by Priyanshu Mishra
4 years, 1 month ago

No vote yet
1 vote

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Comments

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Your answer is correct Who said it is wrong?

Ravi Dwivedi - 4 years, 1 month ago

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

Shanthan Kumar - 4 years ago

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Yes it is...correct!!

Ravi Dwivedi - 4 years ago

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