Question Number 80068 by Pratah last updated on 30/Jan/20
Commented by mr W last updated on 30/Jan/20
$${sir}\:{Pratah}: \\ $$$${are}\:{you}\:{student}?\:{teacher}?\:{professor}? \\ $$$${what}'{s}\:{the}\:{source}\:{of}\:{your}\:{questions}? \\ $$
Commented by Pratah last updated on 30/Jan/20
$$\mathrm{student}.\mathrm{plz}\:\mathrm{solution} \\ $$
Commented by Pratah last updated on 30/Jan/20
$$\mathrm{sir}?\:\:\:\mathrm{john}\:\mathrm{santu} \\ $$
Answered by mr W last updated on 30/Jan/20
$${N}_{\mathrm{1}} =\mathrm{2}{n}\:{times}\:\mathrm{1}=\mathrm{11}…\mathrm{1} \\ $$$${N}_{\mathrm{2}} ={n}\:{times}\:\mathrm{2}=\mathrm{22}…\mathrm{2} \\ $$$${N}_{\mathrm{3}} ={n}\:{times}\:\mathrm{3}=\mathrm{33}…\mathrm{3} \\ $$$$ \\ $$$${N}_{\mathrm{1}} =\frac{\overset{\mathrm{2}{n}\:{times}} {\mathrm{99}…\mathrm{9}}}{\mathrm{9}}=\frac{\mathrm{10}^{\mathrm{2}{n}} −\mathrm{1}}{\mathrm{9}} \\ $$$${N}_{\mathrm{2}} =\frac{\mathrm{2}×\overset{{n}\:{times}} {\mathrm{99}…\mathrm{9}}}{\mathrm{9}}=\frac{\mathrm{2}\left(\mathrm{10}^{{n}} −\mathrm{1}\right)}{\mathrm{9}} \\ $$$${N}_{\mathrm{3}} =\frac{\overset{{n}\:{times}} {\mathrm{99}…\mathrm{9}}}{\mathrm{3}}=\frac{\mathrm{10}^{{n}} −\mathrm{1}}{\mathrm{3}} \\ $$$$ \\ $$$${N}_{\mathrm{1}} −{N}_{\mathrm{2}} =\frac{\mathrm{10}^{\mathrm{2}{n}} −\mathrm{1}−\mathrm{2}\left(\mathrm{10}^{{n}} −\mathrm{1}\right)}{\mathrm{9}} \\ $$$$=\frac{\left(\mathrm{10}^{{n}} \right)^{\mathrm{2}} −\mathrm{2}×\mathrm{10}^{{n}} +\mathrm{1}}{\mathrm{9}}=\left(\frac{\mathrm{10}^{{n}} −\mathrm{1}}{\mathrm{3}}\right)^{\mathrm{2}} \\ $$$$\sqrt{{N}_{\mathrm{1}} −{N}_{\mathrm{2}} }=\frac{\mathrm{10}^{{n}} −\mathrm{1}}{\mathrm{3}}={N}_{\mathrm{3}} \\ $$$$\Rightarrow{proved}! \\ $$
Commented by Pratah last updated on 30/Jan/20
$$\mathrm{thanks} \\ $$
Answered by naka3546 last updated on 31/Jan/20
