How to prove that $\sqrt{2+\sqrt{2+\sqrt{2+\dots}} }=2$
2026-03-29 14:58:55.1774796335
Find the value of $\sqrt{2+\sqrt{2+\sqrt{2+\dots}} }$
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Let $\sqrt{2+\sqrt{2+...}}=a$. Then squaring both sides we have $a^2=2+a$. This is a quadratic equation in $a$. Find the solutions of $a$. Then $a=2 $ or $a=-1$. Since $a>0, a=2$.