#### Question

Show that \sqrt{n+\sqrt{n+\sqrt{n+\sqrt{n+\cdots \infty } } } } = \dfrac{1+\sqrt{1+4n} }{2}

Collected in the board: Square Root

Steven Zheng posted 4 days ago

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}

Steven Zheng posted 4 days ago

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