+4609 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
+118608 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....
