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

Respuesta :

LammettHash

Let

[tex]x = 6 + \dfrac1{2 + \dfrac1{6 + \dfrac1{2 + \cdots}}}[/tex]

Then

[tex]x = 6 + \dfrac1{2 + \dfrac1x}[/tex]

and solving for x gives

[tex]x = 6 + \dfrac x{2x + 1}[/tex]

[tex]x = \dfrac{13x + 6}{2x + 1}[/tex]

[tex]x(2x+1) = 13x + 6[/tex]

[tex]2x^2 + x = 13x + 6[/tex]

[tex]2x^2 - 12x - 6 = 0[/tex]

[tex]x^2 - 6x - 3 = 0[/tex]

By the quadratic formula,

[tex]x = 3 \pm 2\sqrt{3}[/tex]

but x must be a positive number, so only the positive square root solution is valid.

[tex]x = 3 + 2\sqrt{3}[/tex]

We identify a = c = 3 and b = 2, so a + b + c = 8.

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