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Find the value of \(6+\frac{1}{2+\frac{1}{6+\frac{1}{2+\frac{1}{6+\cdots}}}}\)  . Your answer will be of the form \(a+b\sqrt{c}$\)  where no factor of c is a square. Find a+b+c .

 

 

This is for Melody.... Just ignore that dollar sign

tertre  Jun 29, 2017
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3+0 Answers

 #1
avatar+89583 
+1

Hi again Tertre :)

 

Let 

 

\(x=6+\frac{1}{2+\frac{1}{6+\frac{1}{2+\frac{1}{6+\cdots}}}}\\ x=6+\frac{1}{2+\frac{1}{x}}\\ x=6+\frac{1}{\frac{2x+1}{x}}\\ x=6+\frac{x}{2x+1}\\ x(2x+1)=6(2x+1)+x\\ 2x^2+x=12x+6+x\\ 2x^2-12x-6=0\\ x^2-6x-3=0\\ x=\frac{6\pm\sqrt{36+12}}{2}\\ x=\frac{6\pm 4\sqrt{3}}{2}\\ x=3\pm 2\sqrt{3}\\ \text{But x cannot be negative so}\\ x=3+2\sqrt{3}\\~\\ 6+\frac{1}{2+\frac{1}{6+\frac{1}{2+\frac{1}{6+\cdots}}}}=3+2\sqrt{3}\\\)

 

a=3, b=2 c=3 so 

a+b+c=8

 

I am wondering why you were asked for that sum?  

Maybe to make a multiple choice question easier to mark ??

Or perhaps there is a quicker way to find that sum....  

Melody  Jun 29, 2017
 #2
avatar+1105 
0

It's correct, thanks! 

tertre  Jun 29, 2017
edited by tertre  Jun 29, 2017
 #3
avatar+89583 
0


I hope I am not just doing your homework for you. :/

Melody  Jun 29, 2017

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