Find the value of \(x = 1 + \cfrac{1}{2 + \cfrac{1}{2 + \cfrac{1}{2 + \cfrac{1}{2 + \ddots}}}}.\)
+128408 Add 1 to both sides and we have that
1
x + 1 = 2 + __________
2 + 1
________
2 + .......
So we have that
x = 1 + 1
______ multiply through by x + 1
x + 1
x(x + 1) = (x + 1) + 1
x^2 + 1x = 1x + 2 subtract 1x from both sides
x^2 = 2 take the positive square root.....since the original right side is positive
x = √2
+6248 \(\text{let }a = \dfrac{1}{2+\dfrac{1}{2+\dfrac{1}{2+\ddots}}}\)
\(a = \dfrac{1}{2+a}\\ a^2+2a-1=0\\ a = \dfrac{-2\pm \sqrt{4+4}}{2} = -1 \pm \sqrt{2}\\ \text{we can rule out the negative solution since everything is positive}\\ a = -1 + \sqrt{2}\)
\(x= 1+a = \sqrt{2}\)
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+26367 Find the value of
\(x = 1 + \cfrac{1}{2 + \cfrac{1}{2 + \cfrac{1}{2 + \cfrac{1}{2 + \ddots}}}}.\)
\(\begin{array}{|rcll|} \hline \mathbf{x} &=& \mathbf{1 + \cfrac{1}{2 + \cfrac{1}{2 + \cfrac{1}{2 + \cfrac{1}{2 + \ddots}}}} } \\\\ x-1 &=& \cfrac{1}{2 + \cfrac{1}{2 + \cfrac{1}{2 + \cfrac{1}{2 + \ddots}}}} \\\\ \dfrac{1}{x-1} &=& 2+ \cfrac{1}{2 + \cfrac{1}{2 + \cfrac{1}{2 + \ddots}}} \\\\ \dfrac{1}{x-1} &=& 2+ (x-1) \\\\ 1 &=& (x-1)(2+x-1) \\ 1 &=& (x-1)(x+1) \\ 1 &=& x^2-1 \\ x^2 &=& 2 \\ \mathbf{x} &=& \mathbf{\sqrt{2}} \\ \hline \end{array} \)
