x-1-5-11-19-x-

Question Number 85557 by naka3546 last updated on 23/Mar/20

$${x}\:\:=\:\:\sqrt{\mathrm{1}+\:\sqrt{\mathrm{5}+\:\sqrt{\mathrm{11}+\:\sqrt{\mathrm{19}+…}}}} \\ $$$${x}\:\:=\:\:\:? \\ $$

Commented by MJS last updated on 23/Mar/20

$$\mathrm{2}? \\ $$

Commented by naka3546 last updated on 23/Mar/20

$${show}\:\:{your}\:\:{workings}\:,\:{please}\:. \\ $$

Commented by mr W last updated on 23/Mar/20

$${n}=\sqrt{{n}^{\mathrm{2}} }=\sqrt{\left[{n}^{\mathrm{2}} −\left({n}+\mathrm{1}\right)\right]+\left({n}+\mathrm{1}\right)} \\ $$$$=\sqrt{\left[{n}^{\mathrm{2}} −\left({n}+\mathrm{1}\right)\right]+\sqrt{\left[\left({n}+\mathrm{1}\right)^{\mathrm{2}} −\left({n}+\mathrm{2}\right)\right]+\left({n}+\mathrm{2}\right)}} \\ $$$$=\sqrt{\left[{n}^{\mathrm{2}} −\left({n}+\mathrm{1}\right)\right]+\sqrt{\left[\left({n}+\mathrm{1}\right)^{\mathrm{2}} −\left({n}+\mathrm{2}\right)\right]+\sqrt{\left[\left({n}+\mathrm{2}\right)^{\mathrm{2}} −\left({n}+\mathrm{3}\right)\right]+\left({n}+\mathrm{3}\right)}}} \\ $$$$=\sqrt{\left[{n}^{\mathrm{2}} −\left({n}+\mathrm{1}\right)\right]+\sqrt{\left[\left({n}+\mathrm{1}\right)^{\mathrm{2}} −\left({n}+\mathrm{2}\right)\right]+\sqrt{\left[\left({n}+\mathrm{2}\right)^{\mathrm{2}} −\left({n}+\mathrm{3}\right)\right]+\sqrt{\left[\left({n}+\mathrm{3}\right)^{\mathrm{2}} −\left({n}+\mathrm{4}\right)\right]+\sqrt{…}}}}} \\ $$$$ \\ $$$$\Rightarrow{n}=\sqrt{\left[{n}^{\mathrm{2}} −\left({n}+\mathrm{1}\right)\right]+\sqrt{\left[\left({n}+\mathrm{1}\right)^{\mathrm{2}} −\left({n}+\mathrm{2}\right)\right]+\sqrt{\left[\left({n}+\mathrm{2}\right)^{\mathrm{2}} −\left({n}+\mathrm{3}\right)\right]+\sqrt{\left[\left({n}+\mathrm{3}\right)^{\mathrm{2}} −\left({n}+\mathrm{4}\right)\right]+\sqrt{…}}}}} \\ $$$${let}\:{n}=\mathrm{2}: \\ $$$$\mathrm{2}=\sqrt{\mathrm{1}+\sqrt{\mathrm{5}+\sqrt{\mathrm{11}+\sqrt{\mathrm{19}+\sqrt{…}}}}} \\ $$

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