A problem by Anish Shah

Let... $x=\dfrac{6}{1+\dfrac{6}{1+\dfrac{6}{1+\dfrac{6}{1+\dots\infty}}}}$

$\Longrightarrow x=\dfrac{6}{1+x}$

$\Longrightarrow x(x+1)=6=2\times 3=2\times (2+1)$

$\therefore ~~~ x=\fbox{2}$

Note : The last equation is quadratic and has two solutions (the other is $-3$ ), but the question asks for positive value so I didn't go that way... :3

Let $x = \dfrac{6}{1+\dfrac{6}{1 + ... \infty}}$ . Then:

$x = \dfrac{6}{1 + x}$

$x^{2} + x - 6 = 0$

$(x + 3)(x - 2) = 0$

Since only the positive value is required, $x = \boxed{2}$

x = 6/1+x ---> x(x+1)= 6 ---> x^2 + x - 6 = 0 (x-2)(x+3) = 6 ---> x = 2 or -3(impossible). So, x is 2. Answer ; 2

Let $\frac{6}{1+\frac{6}{1+\frac{6}{1+\frac{6}{1+...}}}}=x$ . Therefore, we can dramatically simplify this as $\frac{6}{1+x}=x$ . This leads to $x^2+x=6$ , or $x^2+x-6=0$ . Factoring gives us $(x+3)(x-2)=0$ , so the positive value of the expression is $\boxed{2}$ .

Let, Y= 6/ 1+6/(1+6/1+..infinity )
so, we can write Y =6/(1+6/Y) as Y given , solving the quadratic equation we will get

                             ( Y+3) ( Y -2) =0 .

as we have to take positive number Y = 2

The key to this question lies inside the expression itself. It we let the expression equal $x$ :

$x = \frac {6}{1+\frac {6}{1+\frac {6}{1+\ldots}}}$

Then the value (that I marked with a box) is ALSO equal to x, since it's the same thing!

$x = \frac {6}{1+\boxed{\frac {6}{1+\frac {6}{1+\ldots}}}}$

So instead of that box, let's rewrite it as x. The reason this is important is that we now reduced the fraction that goes on until infinity, to a simple equation that we know how to solve!

$x = \frac {6}{1+x}$

Let's multiply both sides by $1+x$ :

$x(1+x)=6$

Now we open the parenthesis:

$x+x^2=6$

And we have a quadratic equation!

$x^2+x-6=0$

$(x-2)(x+3)=0$

The solutions are therefore $x=2$ OR $x=-3$ ...

Oh wait, we have two solutions?! Remember that the question says:

Find the positive value

So the only answer left is $\boxed{x=2}$ .

$\frac{6}{1+6/1...}$ = x, → $\frac{6}{1+x}$ = x

$x^2+x -6$ = $(x+3)$ $(x-2)$ = 0

$x=-3$ or $x=2$ the question asks the positive value, so $x=$ $\boxed{2}$

Let $f(x) = \frac{6}{1+x}$ and consider the sequence $x, f(x), f^2(x), \cdots$ . Suppose the sequence converges to $z$ for some initial $x$ and $f$ is continuous there, then continuity preserves limits, and we know that $f(z) = z$ is a stable fixed point. Therefore, the answer comes from the solution of the quadratic equation $z = \frac{6}{1+z}$ , which is finitely equivalent to the quadratic $z^2 + z - 6 = 0$ . The positive root of this is $\boxed{2}$ , which is our answer.