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I know that the answer is three and is easy to prove that like this: $$3=\sqrt{1+8}$$ $$ 3=\sqrt{1+2\sqrt{1+15}}$$ $$ 3=\sqrt{1+2\sqrt{1+3\sqrt{1+24}}}$$ $$...$$ but I wanted to know if there is any direct way of calculating the answer.

thanks in advance.

Vfp1649
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  • Similar older questions: https://math.stackexchange.com/q/7204, https://math.stackexchange.com/q/1359873, and some more ... – Martin R Nov 26 '20 at 10:15
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    Your "proof" is actually not rigorous. In the same way, we can "prove" that $3 = \sqrt{1 + \sqrt{1 + \sqrt{1 + \dots}}}$, as $3 = \sqrt{1 + 8} = \sqrt{1 + \sqrt{1 + 63}} = \sqrt{1 + \sqrt{1 + \sqrt{1 + 3968}}} = \dots$. – WhatsUp Nov 26 '20 at 10:17
  • https://youtu.be/leFep9yt3JY –  Nov 26 '20 at 14:18
  • Similar to comment above, your method would show that $4=\sqrt{1+15}=\sqrt{1+2\sqrt{1+221/4}}=\sqrt{1+2\sqrt{1+3\sqrt{1+48697/144}}}=\dots$, or basically any other number you choose... so clearly this approach does not work by itself. – Sil Nov 26 '20 at 15:17

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