Question Number 130744 by rs4089 last updated on 28/Jan/21
Answered by Dwaipayan Shikari last updated on 28/Jan/21
$${n}+\mathrm{1}=\sqrt{{n}^{\mathrm{2}} +\mathrm{2}{n}+\mathrm{1}}=\sqrt{\mathrm{1}+{n}\left({n}+\mathrm{2}\right)} \\ $$$$=\sqrt{\mathrm{1}+{n}\sqrt{\mathrm{1}+\left({n}+\mathrm{1}\right)\left({n}+\mathrm{3}\right)}}=\sqrt{\mathrm{1}+{n}\sqrt{\mathrm{1}+\left({n}+\mathrm{1}\right)\sqrt{\mathrm{1}+\left({n}+\mathrm{2}\right)\sqrt{\mathrm{1}+…}}}} \\ $$$${n}=\mathrm{2} \\ $$$$\mathrm{3}=\sqrt{\mathrm{1}+\mathrm{2}\sqrt{\mathrm{1}+\mathrm{3}\sqrt{\mathrm{1}+\mathrm{4}\sqrt{\mathrm{1}+\mathrm{5}+…}}}} \\ $$$${n}=\mathrm{7} \\ $$$$\mathrm{8}=\sqrt{\mathrm{1}+\mathrm{7}\sqrt{\mathrm{1}+\mathrm{8}\sqrt{\mathrm{1}+\mathrm{9}\sqrt{\mathrm{1}+\mathrm{10}+…}}}} \\ $$