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Let $E(n) = \dfrac{4^n}{4^n+2}$.

If the value of $E(\frac{1}{2018})+E(\frac{2}{2018})+\dots+E(\frac{2017}{2018})=\frac{a}{b}$ (in lowest terms), find $b$.

I tried solving this question using telescopic sum but was unable to solve

JMoravitz
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    Here's a MathJax tutorial :) – Shaun Jan 07 '20 at 13:52
  • I have typeset your image... please do it yourself from now on using the information provided in Shaun's link. The original image you posted was an eyesore... far too big and sideways. Using images instead of text is highly frowned upon for a number of reasons, among which being that having text instead helps the system identify possible duplicates and related problems, allows users with poor/no vision to use tools to still hear the question, and makes it generally easier to read and understand. – JMoravitz Jan 07 '20 at 14:11
  • Hint: try smaller sums. Replace the denominator by a smaller value. – lulu Jan 07 '20 at 14:14

1 Answers1

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Let $\displaystyle f(x)=\frac{4^x}{4^x+2},$ Then $\displaystyle f(1-x)=\frac{4^{1-x}}{4^{1-x}+2}=\frac{2}{4^x+2}$

So $$f(x)+f(1-x)=\frac{4^x+2}{4^x+2}=1.$$

So Put $\displaystyle x=\frac{1}{2018},\frac{2}{2018},\frac{3}{2018},\cdots \cdots ,\frac{1008}{2018}$

So $$f\bigg(\frac{1}{2018}\bigg)+f\bigg(\frac{2}{2018}\bigg)+f\bigg(\frac{3}{2018}\bigg)+\cdots +\cdots +f\bigg(\frac{1009}{2018}\bigg)+f\bigg(\frac{1010}{2018}\bigg)\cdots +f\bigg(\frac{2017}{2018}\bigg)=1008+\frac{1}{2}$$

For calculation of $\displaystyle f\bigg(\frac{1009}{2018}\bigg),$

Put $\displaystyle x=\frac{1009}{2018}$ in $f(x)+f(1-x)=1,$

Getting $$\displaystyle f\bigg(\frac{1009}{2018}\bigg)=\frac{1}{2}.$$

Thomas Lesgourgues
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jacky
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