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Question Number 71 by mike last updated on 25/Jan/15

(√(1+(√(1+(√(1+…))))))=?

$$\sqrt{\mathrm{1}+\sqrt{\mathrm{1}+\sqrt{\mathrm{1}+\ldots}}}=? \\ $$

Answered by jayant last updated on 17/Nov/14

1

$$\mathrm{1} \\ $$

Commented by 123456 last updated on 13/Dec/14

+0,618...  aproximadamente  pudim→rosquinha

$$+\mathrm{0},\mathrm{618}... \\ $$$${aproximadamente} \\ $$$${pudim}\rightarrow{rosquinha} \\ $$

Commented by prakash jain last updated on 13/Dec/14

It should be ≈1.618

$$\mathrm{It}\:\mathrm{should}\:\mathrm{be}\:\approx\mathrm{1}.\mathrm{618} \\ $$

Answered by newuser last updated on 17/Nov/14

(√(1+(√(1+(√(1+…))))))=x  ⇒(√(1+x))=x  ⇒1+x=x^2   ⇒x^2 −x−1=0  ⇒x=((1±(√(1+4)))/2)=((1±(√5))/2)  taking positive solution x=((1+(√5))/2)  (√(1+(√(1+(√(1+…))))))=((1+(√5))/2)

$$\sqrt{\mathrm{1}+\sqrt{\mathrm{1}+\sqrt{\mathrm{1}+\ldots}}}={x} \\ $$$$\Rightarrow\sqrt{\mathrm{1}+{x}}={x} \\ $$$$\Rightarrow\mathrm{1}+{x}={x}^{\mathrm{2}} \\ $$$$\Rightarrow{x}^{\mathrm{2}} −{x}−\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow{x}=\frac{\mathrm{1}\pm\sqrt{\mathrm{1}+\mathrm{4}}}{\mathrm{2}}=\frac{\mathrm{1}\pm\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$$$\mathrm{taking}\:\mathrm{positive}\:\mathrm{solution}\:{x}=\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$$$\sqrt{\mathrm{1}+\sqrt{\mathrm{1}+\sqrt{\mathrm{1}+\ldots}}}=\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$

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