Question Number 154034 by amin96 last updated on 13/Sep/21
$${prove}\:{that}\: \\ $$$${n}+{x}=\sqrt{{n}^{\mathrm{2}} +{x}\sqrt{{n}^{\mathrm{2}} +\left({x}+{n}\right)\sqrt{{n}^{\mathrm{2}} +\left({x}+\mathrm{2}{n}\right)\sqrt{{n}^{\mathrm{2}} \ldots}}}} \\ $$$${Ramanujan}'{s}\:{nested}\:{radikal} \\ $$
Answered by mr W last updated on 13/Sep/21
$$\left({n}+{x}\right)^{\mathrm{2}} ={n}^{\mathrm{2}} +\mathrm{2}{xn}+{x}^{\mathrm{2}} \\ $$$$\left({n}+{x}\right)^{\mathrm{2}} ={n}^{\mathrm{2}} +{x}\left({n}+{x}+{n}\right)\:\:...\left({i}\right) \\ $$$$\Rightarrow{n}+{x}=\sqrt{{n}^{\mathrm{2}} +{x}\left({n}+{x}+{n}\right)}\:\:...\left({ii}\right) \\ $$$${in}\:\left({i}\right)\:{replace}\:{x}\:{with}\:{x}+{n}, \\ $$$$\left({n}+{x}+{n}\right)^{\mathrm{2}} ={n}^{\mathrm{2}} +\left({x}+{n}\right)\left({n}+{x}+\mathrm{2}{n}\right) \\ $$$$\Rightarrow\left({n}+{x}+{n}\right)=\sqrt{{n}^{\mathrm{2}} +\left({x}+{n}\right)\left({n}+{x}+\mathrm{2}{n}\right)} \\ $$$${put}\:{this}\:{into}\:\left({ii}\right), \\ $$$$\Rightarrow{n}+{x}=\sqrt{{n}^{\mathrm{2}} +{x}\sqrt{{n}^{\mathrm{2}} +\left({x}+{n}\right)\left({n}+{x}+\mathrm{2}{n}\right)}}\:\:\:...\left({iii}\right) \\ $$$${in}\:\left({i}\right)\:{replace}\:{x}\:{with}\:{x}+\mathrm{2}{n}, \\ $$$$\left({n}+{x}+\mathrm{2}{n}\right)^{\mathrm{2}} ={n}^{\mathrm{2}} +\left({x}+\mathrm{2}{n}\right)\left({n}+{x}+\mathrm{3}{n}\right) \\ $$$$\Rightarrow\left({n}+{x}+\mathrm{2}{n}\right)=\sqrt{{n}^{\mathrm{2}} +\left({x}+\mathrm{2}{n}\right)\left({n}+{x}+\mathrm{3}{n}\right)} \\ $$$${put}\:{this}\:{into}\:\left({iii}\right), \\ $$$$\Rightarrow{n}+{x}=\sqrt{{n}^{\mathrm{2}} +{x}\sqrt{{n}^{\mathrm{2}} +\left({x}+{n}\right)\sqrt{{n}^{\mathrm{2}} +\left({x}+\mathrm{2}{n}\right)\left({n}+{x}+\mathrm{3}{n}\right)}}}\:\:\:...\left({iv}\right) \\ $$$${and}\:{so}\:{on}\:{so}\:{on}\:{so}\:{on}.... \\ $$
Commented by amin96 last updated on 13/Sep/21
$${good}\:{work}\:{sir} \\ $$