Question

Find the value of
`$6+(1)/(2+(1)/(6+(1)/(2+(1)/(6+\cdots))))$.` Your answer will be of the form
`$a+b√(c)$` where no factor of $c$ (other than $1$) is a square. Find $a+b+c$.

Answer 1

Let

`x = 6 + \frac1{2 + \frac1{6 + \frac1{2 + \cdots}}}`

Then

`x = 6 + \frac1{2 + \frac1x}`

and solving for x gives

`x = 6 + \frac x{2x + 1}`

`x = (13x + 6)/(2x + 1)`

`x(2x+1) = 13x + 6`

`2x^2 + x = 13x + 6`

`2x^2 - 12x - 6 = 0`

`x^2 - 6x - 3 = 0`

By the quadratic formula,

`x = 3 \pm 2√(3)`

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

`x = 3 + 2√(3)`

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