Sigma

Algebra Level 3

If f ( x ) = 4 x 4 x + 2 f(x)=\dfrac{4^{x}}{4^{x}+2} .
Compute the value of ∑ n = 1 1000 f ( n 1000 ) \displaystyle \sum_{n=1}^{1000} f\left(\dfrac{n}{1000}\right) .

Give your answer to 2 decimal places.


The answer is 500.17.

Ayush G Rai by Ayush G Rai , 20, India

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1 solution

Sabhrant Sachan
May 2, 2016

We have f ( x ) = 4 x 4 x + 2 Now f ( 1 − x ) = 4 1 − x 4 1 − x + 2 ⟹ f ( 1 − x ) = 2 2 + 4 x f ( x ) + f ( 1 − x ) = 4 x + 2 2 + 4 x = 1 Let S be the required Sum S = f ( 1 1000 ) + f ( 2 1000 ) + ⋯ + f ( 1000 1000 ) S − f ( 1 ) = f ( 999 1000 ) + f ( 998 1000 ) + ⋯ + f ( 1 1000 ) Add Both the equations, we get 2 S − f ( 1 ) = [ f ( 1 1000 ) + f ( 999 1000 ) ] + [ f ( 2 1000 ) + f ( 998 1000 ) ] + . . . + [ f ( 999 1000 ) + f ( 1 1000 ) ] ⏟ 999 times + f ( 1 ) ⟹ 2 S − f ( 1 ) = 999 + f ( 1 ) ⟹ S = 999 2 + f ( 1 ) ⟹ S = 499.5 + 0.667 ⟹ 500.17 \text {We have } f(x)=\dfrac{4^x}{4^x+2} \\ \text {Now } f(1-x)=\dfrac{4^{1-x}}{4^{1-x}+2} \\ \implies f(1-x)=\dfrac2{2+4^x} \\ f(x)+f(1-x)=\dfrac{4^x+2}{2+4^x}=1 \\ \text{Let } \color{#3D99F6}{S} \text{ be the required Sum} \\ S=f(\dfrac1{1000})+f(\dfrac2{1000})+\cdots+f(\dfrac{1000}{1000})\\ S-f(1)=f(\dfrac{999}{1000})+f(\dfrac{998}{1000})+\cdots+f(\dfrac1{1000})\\ \text {Add Both the equations, we get } \\ 2S-f(1)=\underbrace{[f(\dfrac{1}{1000})+f(\dfrac{999}{1000})]+[f(\dfrac{2}{1000})+f(\dfrac{998}{1000})]+...+[f(\dfrac{999}{1000})+f(\dfrac{1}{1000})]}_{\color{#D61F06}{\text {999 times}}}+f(1)\\ \implies 2S-f(1)=999+f(1)\\ \implies S=\dfrac{999}2+f(1)\\ \implies S=499.5+0.667 \implies \color{#3D99F6}{500.17}

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