Question Number 71 by mike last updated on 25/Jan/15
$$\sqrt{\mathrm{1}+\sqrt{\mathrm{1}+\sqrt{\mathrm{1}+\ldots}}}=? \\ $$
Answered by jayant last updated on 17/Nov/14
$$\mathrm{1} \\ $$
Commented by 123456 last updated on 13/Dec/14
$$+\mathrm{0},\mathrm{618}… \\ $$$${aproximadamente} \\ $$$${pudim}\rightarrow{rosquinha} \\ $$
Commented by prakash jain last updated on 13/Dec/14
$$\mathrm{It}\:\mathrm{should}\:\mathrm{be}\:\approx\mathrm{1}.\mathrm{618} \\ $$
Answered by newuser last updated on 17/Nov/14
$$\sqrt{\mathrm{1}+\sqrt{\mathrm{1}+\sqrt{\mathrm{1}+\ldots}}}={x} \\ $$$$\Rightarrow\sqrt{\mathrm{1}+{x}}={x} \\ $$$$\Rightarrow\mathrm{1}+{x}={x}^{\mathrm{2}} \\ $$$$\Rightarrow{x}^{\mathrm{2}} −{x}−\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow{x}=\frac{\mathrm{1}\pm\sqrt{\mathrm{1}+\mathrm{4}}}{\mathrm{2}}=\frac{\mathrm{1}\pm\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$$$\mathrm{taking}\:\mathrm{positive}\:\mathrm{solution}\:{x}=\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$$$\sqrt{\mathrm{1}+\sqrt{\mathrm{1}+\sqrt{\mathrm{1}+\ldots}}}=\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$