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Question Number 17991 by alex041103 last updated on 13/Jul/17

Evaluate (√(1+2(√(1+3(√(1+4(√(1+...))))))))

$${Evaluate}\:\sqrt{\mathrm{1}+\mathrm{2}\sqrt{\mathrm{1}+\mathrm{3}\sqrt{\mathrm{1}+\mathrm{4}\sqrt{\mathrm{1}+...}}}} \\ $$

Commented by alex041103 last updated on 13/Jul/17

just for fun whoever wants can try to generalize (√(A+B_1 (√(A+B_2 (√(A+B_3 (√(A+ ...))))))))

$${just}\:{for}\:{fun} \\ $$$${whoever}\:{wants}\:{can}\:{try}\:{to}\:{generalize} \\ $$$$\sqrt{{A}+{B}_{\mathrm{1}} \sqrt{{A}+{B}_{\mathrm{2}} \sqrt{{A}+{B}_{\mathrm{3}} \sqrt{{A}+\:...}}}} \\ $$

Commented by alex041103 last updated on 13/Jul/17

Answers to Q.17989 and Q.17991 on 15^(th) July

$${Answers}\:{to}\:{Q}.\mathrm{17989}\:{and}\:{Q}.\mathrm{17991}\:{on} \\ $$$$\mathrm{15}^{{th}} \:{July} \\ $$

Commented by prakash jain last updated on 13/Jul/17

3=(√(1+8)) =(√(1+2∙4)) =(√(1+2(√(16)))) =(√(1+2(√(1+15)))) =(√(1+2(√(1+3.5)))) =(√(1+2(√(1+3.(√(1+24)))))) =(√(1+2(√(1+3.(√(1+4.6)))))) =(√(1+2(√(1+3.(√(1+4.(√(1+5.(√(1+6(√(1+))..))))))))))

$$\mathrm{3}=\sqrt{\mathrm{1}+\mathrm{8}} \\ $$$$=\sqrt{\mathrm{1}+\mathrm{2}\centerdot\mathrm{4}} \\ $$$$=\sqrt{\mathrm{1}+\mathrm{2}\sqrt{\mathrm{16}}} \\ $$$$=\sqrt{\mathrm{1}+\mathrm{2}\sqrt{\mathrm{1}+\mathrm{15}}} \\ $$$$=\sqrt{\mathrm{1}+\mathrm{2}\sqrt{\mathrm{1}+\mathrm{3}.\mathrm{5}}} \\ $$$$=\sqrt{\mathrm{1}+\mathrm{2}\sqrt{\mathrm{1}+\mathrm{3}.\sqrt{\mathrm{1}+\mathrm{24}}}} \\ $$$$=\sqrt{\mathrm{1}+\mathrm{2}\sqrt{\mathrm{1}+\mathrm{3}.\sqrt{\mathrm{1}+\mathrm{4}.\mathrm{6}}}} \\ $$$$=\sqrt{\mathrm{1}+\mathrm{2}\sqrt{\mathrm{1}+\mathrm{3}.\sqrt{\mathrm{1}+\mathrm{4}.\sqrt{\mathrm{1}+\mathrm{5}.\sqrt{\mathrm{1}+\mathrm{6}\sqrt{\mathrm{1}+}..}}}}} \\ $$

Answered by prakash jain last updated on 13/Jul/17

3

$$\mathrm{3} \\ $$

Commented by alex041103 last updated on 13/Jul/17

Bonus: This infinite beauty was discovered by the genius indian mathmatition Srinivasa Ramanudjan

$${Bonus}:\:{This}\:{infinite}\:{beauty}\:{was}\:{discovered} \\ $$$${by}\:{the}\:{genius}\:{indian}\:{mathmatition} \\ $$$${Srinivasa}\:{Ramanudjan} \\ $$

Answered by alex041103 last updated on 14/Jul/17

Generalization let A =x^2 and B_(n+1) =x+B_n B_1 =(√B_1 ^2 )=(√(A+(B_1 ^2 −A)))= =(√(A+(B_1 −x)(B_1 +x)))= =(√(A+B_0 (√(A+B_2 ^2 −x^2 ))))= =(√(A+B_0 (√(A+(B_2 −x)(B_2 +x)))))= =(√(A+B_0 (√(A+B_1 (√B_3 ^2 )))))= =....= =(√(A+B_0 (√(A+B_1 (√(A+B_2 (√(A+B_3 (√(...)))))))))) B_1 =(√(A+B_0 (√(A+B_1 (√(A+B_2 (√(A+B_3 (√(...)))))))))) Example: A=1 and B_1 =4 4=(√(1+3(√(1+4(√(1+5(√(1+6(√(...))))))))))

$${Generalization} \\ $$$${let}\:{A}\:={x}^{\mathrm{2}} \:{and}\:{B}_{{n}+\mathrm{1}} ={x}+{B}_{{n}} \\ $$$${B}_{\mathrm{1}} =\sqrt{{B}_{\mathrm{1}} ^{\mathrm{2}} }=\sqrt{{A}+\left({B}_{\mathrm{1}} ^{\mathrm{2}} −{A}\right)}= \\ $$$$=\sqrt{{A}+\left({B}_{\mathrm{1}} −{x}\right)\left({B}_{\mathrm{1}} +{x}\right)}= \\ $$$$=\sqrt{{A}+{B}_{\mathrm{0}} \sqrt{{A}+{B}_{\mathrm{2}} ^{\mathrm{2}} −{x}^{\mathrm{2}} }}= \\ $$$$=\sqrt{{A}+{B}_{\mathrm{0}} \sqrt{{A}+\left({B}_{\mathrm{2}} −{x}\right)\left({B}_{\mathrm{2}} +{x}\right)}}= \\ $$$$=\sqrt{{A}+{B}_{\mathrm{0}} \sqrt{{A}+{B}_{\mathrm{1}} \sqrt{{B}_{\mathrm{3}} ^{\mathrm{2}} }}}= \\ $$$$=....= \\ $$$$=\sqrt{{A}+{B}_{\mathrm{0}} \sqrt{{A}+{B}_{\mathrm{1}} \sqrt{{A}+{B}_{\mathrm{2}} \sqrt{{A}+{B}_{\mathrm{3}} \sqrt{...}}}}} \\ $$$${B}_{\mathrm{1}} =\sqrt{{A}+{B}_{\mathrm{0}} \sqrt{{A}+{B}_{\mathrm{1}} \sqrt{{A}+{B}_{\mathrm{2}} \sqrt{{A}+{B}_{\mathrm{3}} \sqrt{...}}}}} \\ $$$$ \\ $$$${Example}: \\ $$$${A}=\mathrm{1}\:{and}\:{B}_{\mathrm{1}} =\mathrm{4} \\ $$$$\mathrm{4}=\sqrt{\mathrm{1}+\mathrm{3}\sqrt{\mathrm{1}+\mathrm{4}\sqrt{\mathrm{1}+\mathrm{5}\sqrt{\mathrm{1}+\mathrm{6}\sqrt{...}}}}} \\ $$

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