Proof that $\sqrt{1+2\sqrt{1+2\sqrt{1+2\sqrt{1+2(\ldots)}}}} = 1+\sqrt{2}$

Question

I saw this problem at the 2017 math counts competition and one of the kids solved it in 5 seconds. I played around with is seeing that it could be represented as infinite nested function $f(x)=\sqrt{1+2x}$ but couldn't get much further.

https://youtu.be/vFTeN17Z4rc?t=47m54s

Answer 1

Your nested function should be $f(x)=\sqrt{1+2f(x)}$

So as BAI suggested, solve $y=\sqrt{1+2y}$

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Hint: square both sides and find the positive solution to the quadratic.   (Why only the positive?)

Answer 2

Yes, of course!

Since $\sqrt{3+2\sqrt2}=1+\sqrt2,$ we obtain: $$...=\sqrt{1+2\sqrt{1+2\sqrt{1+2(1+\sqrt2)}}}=\sqrt{1+2\sqrt{1+2(1+\sqrt2)}}=$$ $$=\sqrt{1+2(1+\sqrt2)}=1+\sqrt2.$$ We can see it during one second.

Answer 3

Note: $$1+\sqrt{2}=\sqrt{(1+\sqrt{2})^2}=\sqrt{1+2(1+\sqrt{2})}=...$$

Answer 4

Assume $\sqrt{1+2\sqrt{1+2\sqrt{\dots}}}=x$.

Putting this from the second root in the equation, we get $\sqrt{1+2x}=1+\sqrt{2}$ solving this $x=1+\sqrt{2}$. so $$\sqrt{1+2\sqrt{1+2\sqrt{\dots}}}=1+\sqrt{2}$$