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Question Number 110204 by Algoritm last updated on 27/Aug/20

Commented by Algoritm last updated on 27/Aug/20

⧫ prove that !

$$\blacklozenge\:\boldsymbol{\mathrm{prove}}\:\boldsymbol{\mathrm{that}}\:! \\ $$

Commented by prakash jain last updated on 28/Aug/20

(2^n ∙4+1)^2 =4^(n+2) +2^(n+1) ∙4+1 (2^n .4+1)=(√(4^(n+2) +(2^(n+1) ∙4+1))) 3=(√9) =(√(4+(2^0 .4+1))) =(√(4+(√(4^2 +2^1 .4+1)))) =(√(4+(√(4^2 +(√(4^3 +2^2 ∙4+1)))))) and so on

$$\left(\mathrm{2}^{{n}} \centerdot\mathrm{4}+\mathrm{1}\right)^{\mathrm{2}} =\mathrm{4}^{{n}+\mathrm{2}} +\mathrm{2}^{{n}+\mathrm{1}} \centerdot\mathrm{4}+\mathrm{1} \\ $$$$\left(\mathrm{2}^{{n}} .\mathrm{4}+\mathrm{1}\right)=\sqrt{\mathrm{4}^{{n}+\mathrm{2}} +\left(\mathrm{2}^{{n}+\mathrm{1}} \centerdot\mathrm{4}+\mathrm{1}\right)} \\ $$$$\mathrm{3}=\sqrt{\mathrm{9}} \\ $$$$=\sqrt{\mathrm{4}+\left(\mathrm{2}^{\mathrm{0}} .\mathrm{4}+\mathrm{1}\right)} \\ $$$$=\sqrt{\mathrm{4}+\sqrt{\mathrm{4}^{\mathrm{2}} +\mathrm{2}^{\mathrm{1}} .\mathrm{4}+\mathrm{1}}} \\ $$$$=\sqrt{\mathrm{4}+\sqrt{\mathrm{4}^{\mathrm{2}} +\sqrt{\mathrm{4}^{\mathrm{3}} +\mathrm{2}^{\mathrm{2}} \centerdot\mathrm{4}+\mathrm{1}}}} \\ $$$$\mathrm{and}\:\mathrm{so}\:\mathrm{on} \\ $$

Commented by Algoritm last updated on 27/Aug/20

sir (√9)=(√(4+(2×4+1)))=(√(4+9)) ???

$$\mathrm{sir}\:\:\sqrt{\mathrm{9}}=\sqrt{\mathrm{4}+\left(\mathrm{2}×\mathrm{4}+\mathrm{1}\right)}=\sqrt{\mathrm{4}+\mathrm{9}}\:\:??? \\ $$

Commented by prakash jain last updated on 28/Aug/20

correct. but the basic idea remains the same.

$$\mathrm{correct}.\:\mathrm{but}\:\mathrm{the}\:\mathrm{basic}\:\mathrm{idea}\:\mathrm{remains} \\ $$$$\mathrm{the}\:\mathrm{same}. \\ $$

Answered by Dwaipayan Shikari last updated on 27/Aug/20

3=(√(4+(√(4^2 +9)))) 3=(√(4+(√(4^2 +(√(4^3 +17)))))) 3=(√(4+(√(4^2 +(√(4^3 +(√(4^4 +33)))))))) 3=(√(4+(√(4^2 +(√(4^3 +(√(4^4 +(√(4^5 +65)))))))))) 3=(√(4+(√(4^2 +(√(4^3 +(√(4^4 +(√(4^5 +......))))))))))

$$\mathrm{3}=\sqrt{\mathrm{4}+\sqrt{\mathrm{4}^{\mathrm{2}} +\mathrm{9}}} \\ $$$$\mathrm{3}=\sqrt{\mathrm{4}+\sqrt{\mathrm{4}^{\mathrm{2}} +\sqrt{\mathrm{4}^{\mathrm{3}} +\mathrm{17}}}} \\ $$$$\mathrm{3}=\sqrt{\mathrm{4}+\sqrt{\mathrm{4}^{\mathrm{2}} +\sqrt{\mathrm{4}^{\mathrm{3}} +\sqrt{\mathrm{4}^{\mathrm{4}} +\mathrm{33}}}}} \\ $$$$\mathrm{3}=\sqrt{\mathrm{4}+\sqrt{\mathrm{4}^{\mathrm{2}} +\sqrt{\mathrm{4}^{\mathrm{3}} +\sqrt{\mathrm{4}^{\mathrm{4}} +\sqrt{\mathrm{4}^{\mathrm{5}} +\mathrm{65}}}}}} \\ $$$$\mathrm{3}=\sqrt{\mathrm{4}+\sqrt{\mathrm{4}^{\mathrm{2}} +\sqrt{\mathrm{4}^{\mathrm{3}} +\sqrt{\mathrm{4}^{\mathrm{4}} +\sqrt{\mathrm{4}^{\mathrm{5}} +......}}}}} \\ $$

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