Question Number 17991 by alex041103 last updated on 13/Jul/17
$${Evaluate}\:\sqrt{\mathrm{1}+\mathrm{2}\sqrt{\mathrm{1}+\mathrm{3}\sqrt{\mathrm{1}+\mathrm{4}\sqrt{\mathrm{1}+…}}}} \\ $$
Commented by alex041103 last updated on 13/Jul/17
$${just}\:{for}\:{fun} \\ $$$${whoever}\:{wants}\:{can}\:{try}\:{to}\:{generalize} \\ $$$$\sqrt{{A}+{B}_{\mathrm{1}} \sqrt{{A}+{B}_{\mathrm{2}} \sqrt{{A}+{B}_{\mathrm{3}} \sqrt{{A}+\:…}}}} \\ $$
Commented by alex041103 last updated on 13/Jul/17
$${Answers}\:{to}\:{Q}.\mathrm{17989}\:{and}\:{Q}.\mathrm{17991}\:{on} \\ $$$$\mathrm{15}^{{th}} \:{July} \\ $$
Commented by prakash jain last updated on 13/Jul/17
$$\mathrm{3}=\sqrt{\mathrm{1}+\mathrm{8}} \\ $$$$=\sqrt{\mathrm{1}+\mathrm{2}\centerdot\mathrm{4}} \\ $$$$=\sqrt{\mathrm{1}+\mathrm{2}\sqrt{\mathrm{16}}} \\ $$$$=\sqrt{\mathrm{1}+\mathrm{2}\sqrt{\mathrm{1}+\mathrm{15}}} \\ $$$$=\sqrt{\mathrm{1}+\mathrm{2}\sqrt{\mathrm{1}+\mathrm{3}.\mathrm{5}}} \\ $$$$=\sqrt{\mathrm{1}+\mathrm{2}\sqrt{\mathrm{1}+\mathrm{3}.\sqrt{\mathrm{1}+\mathrm{24}}}} \\ $$$$=\sqrt{\mathrm{1}+\mathrm{2}\sqrt{\mathrm{1}+\mathrm{3}.\sqrt{\mathrm{1}+\mathrm{4}.\mathrm{6}}}} \\ $$$$=\sqrt{\mathrm{1}+\mathrm{2}\sqrt{\mathrm{1}+\mathrm{3}.\sqrt{\mathrm{1}+\mathrm{4}.\sqrt{\mathrm{1}+\mathrm{5}.\sqrt{\mathrm{1}+\mathrm{6}\sqrt{\mathrm{1}+}..}}}}} \\ $$
Answered by prakash jain last updated on 13/Jul/17
$$\mathrm{3} \\ $$
Commented by alex041103 last updated on 13/Jul/17
$${Bonus}:\:{This}\:{infinite}\:{beauty}\:{was}\:{discovered} \\ $$$${by}\:{the}\:{genius}\:{indian}\:{mathmatition} \\ $$$${Srinivasa}\:{Ramanudjan} \\ $$
Answered by alex041103 last updated on 14/Jul/17
$${Generalization} \\ $$$${let}\:{A}\:={x}^{\mathrm{2}} \:{and}\:{B}_{{n}+\mathrm{1}} ={x}+{B}_{{n}} \\ $$$${B}_{\mathrm{1}} =\sqrt{{B}_{\mathrm{1}} ^{\mathrm{2}} }=\sqrt{{A}+\left({B}_{\mathrm{1}} ^{\mathrm{2}} −{A}\right)}= \\ $$$$=\sqrt{{A}+\left({B}_{\mathrm{1}} −{x}\right)\left({B}_{\mathrm{1}} +{x}\right)}= \\ $$$$=\sqrt{{A}+{B}_{\mathrm{0}} \sqrt{{A}+{B}_{\mathrm{2}} ^{\mathrm{2}} −{x}^{\mathrm{2}} }}= \\ $$$$=\sqrt{{A}+{B}_{\mathrm{0}} \sqrt{{A}+\left({B}_{\mathrm{2}} −{x}\right)\left({B}_{\mathrm{2}} +{x}\right)}}= \\ $$$$=\sqrt{{A}+{B}_{\mathrm{0}} \sqrt{{A}+{B}_{\mathrm{1}} \sqrt{{B}_{\mathrm{3}} ^{\mathrm{2}} }}}= \\ $$$$=….= \\ $$$$=\sqrt{{A}+{B}_{\mathrm{0}} \sqrt{{A}+{B}_{\mathrm{1}} \sqrt{{A}+{B}_{\mathrm{2}} \sqrt{{A}+{B}_{\mathrm{3}} \sqrt{…}}}}} \\ $$$${B}_{\mathrm{1}} =\sqrt{{A}+{B}_{\mathrm{0}} \sqrt{{A}+{B}_{\mathrm{1}} \sqrt{{A}+{B}_{\mathrm{2}} \sqrt{{A}+{B}_{\mathrm{3}} \sqrt{…}}}}} \\ $$$$ \\ $$$${Example}: \\ $$$${A}=\mathrm{1}\:{and}\:{B}_{\mathrm{1}} =\mathrm{4} \\ $$$$\mathrm{4}=\sqrt{\mathrm{1}+\mathrm{3}\sqrt{\mathrm{1}+\mathrm{4}\sqrt{\mathrm{1}+\mathrm{5}\sqrt{\mathrm{1}+\mathrm{6}\sqrt{…}}}}} \\ $$