Question
Show that \sqrt{n+\sqrt{n+\sqrt{n+\sqrt{n+\cdots \infty } } } } = \dfrac{1+\sqrt{1+4n} }{2}
Answer
Let
x =\sqrt{n+\sqrt{n+\sqrt{n+\sqrt{n+\cdots \infty } } } }
when the layers of square root go to infinite
which makes no difference starting to count from the second square root.
Then, we get the following equation
x = \sqrt{n+x}
Then,
x > 0
Square both sides results in a quadratic equation
x^2-x-n=0
Using the root formula for a quadratic equation, we get
x = \dfrac{1\pm\sqrt{1^2-4\cdot 1\cdot (-n)} }{2} = \dfrac{1\pm\sqrt{1+4n} }{2}
Cancel the negative solution, then,
x = \dfrac{1+\sqrt{1+4n} }{2}
Therfore,
\sqrt{n+\sqrt{n+\sqrt{n+\sqrt{n+\cdots \infty } } } } = \dfrac{1+\sqrt{1+4n} }{2}