2001 Problem!

Algebra Level 4

f ( 1 2001 ) + f ( 2 2001 ) + f ( 3 2001 ) + ⋯ + f ( 2000 2001 ) f \left( \frac 1{2001} \right) + f \left( \frac 2{2001} \right) + f \left( \frac 3{2001} \right) +\cdots + f \left( \frac {2000}{2001} \right)

Let f ( x ) = 2 4 x + 2 f(x) = \dfrac 2{4^x+2} for real numbers x x . Evaluate the expression above.


The answer is 1000.

Novril Razenda by Novril Razenda , 21, Indonesia

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2 solutions

Abhishek Sinha
Jul 2, 2016

Just note that f ( x ) + f ( 1 − x ) = 1 f(x) + f(1-x)=1 Now pair up the terms.

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As suggested by @Abhishek Sinha :

f ( x ) = 2 4 x + 2 ⟹ f ( 1 − x ) = 2 4 1 − x + 2 = 2 4 4 x + 2 = 4 x 2 + 4 x = 4 x + 2 − 2 4 x + 2 = 1 − f ( x ) ⟹ f ( x ) + f ( 1 − x ) = 1 ⟹ f ( 1 2001 ) + f ( 2000 2001 ) = 1 As 1 − 1 2001 = 2000 2001 f ( 2 2001 ) + f ( 1999 2001 ) = 1 f ( 3 2001 ) + f ( 1998 2001 ) = 1 . . . . . . \begin{aligned} f(x) & = \frac 2{4^x+2} \\ \implies f(1-x) & = \frac 2{4^{1-x}+2} \\ & = \frac 2{\frac 4{4^x}+2} \\ & = \frac {4^x}{2+4^x} \\ & = \frac {4^x+2-2}{4^x+2} \\ & = 1 - f(x) \\ \implies f(x) + f(1-x) & = 1 \\ \implies f \left(\frac 1{2001} \right) + f \left(\color{#3D99F6}{\frac {2000}{2001}} \right) & = 1 & \small \color{#3D99F6}{\text{As } 1-\frac 1{2001} = \frac{2000}{2001}} \\ f \left(\frac 2{2001} \right) + f \left(\frac {1999}{2001} \right) & = 1 \\ f \left(\frac 3{2001} \right) + f \left(\frac {1998}{2001} \right) & = 1 \\ ... & \quad \ ... \end{aligned}

⟹ f ( 1 2001 ) + f ( 2 2001 ) + f ( 3 2001 ) + . . . + f ( 2000 2001 ) = 1000 \implies f \left(\frac 1{2001} \right) + f \left(\frac 2{2001} \right) +f \left(\frac 3{2001} \right) +...+f \left(\frac {2000}{2001} \right) = \boxed{1000}

Chew-Seong Cheong - 4 years, 11 months ago

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Nice method :) +1 :)

Novril Razenda - 4 years, 11 months ago

@Abhishek Sinha Nice +1 :D

Novril Razenda - 4 years, 11 months ago
Shree Ganesh
Jul 3, 2016

I did by summing up first and last term

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It's equally such as f ( x ) + f ( 1 − x ) f(x)+f(1-x) ;)

Novril Razenda - 4 years, 11 months ago

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