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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?

$${sir}\:{Pratah}: \\ $$$${are}\:{you}\:{student}?\:{teacher}?\:{professor}? \\ $$$${what}'{s}\:{the}\:{source}\:{of}\:{your}\:{questions}? \\ $$

Commented by Pratah last updated on 30/Jan/20

student.plz solution

$$\mathrm{student}.\mathrm{plz}\:\mathrm{solution} \\ $$

Commented by Pratah last updated on 30/Jan/20

sir? john santu

$$\mathrm{sir}?\:\:\:\mathrm{john}\:\mathrm{santu} \\ $$

Answered by mr W last updated on 30/Jan/20

N_1 =2n times 1=11...1 N_2 =n times 2=22...2 N_3 =n times 3=33...3 N_1 =((99...9^(2n times) )/9)=((10^(2n) −1)/9) N_2 =((2×99...9^(n times) )/9)=((2(10^n −1))/9) N_3 =((99...9^(n times) )/3)=((10^n −1)/3) N_1 −N_2 =((10^(2n) −1−2(10^n −1))/9) =(((10^n )^2 −2×10^n +1)/9)=(((10^n −1)/3))^2 (√(N_1 −N_2 ))=((10^n −1)/3)=N_3 ⇒proved!

$${N}_{\mathrm{1}} =\mathrm{2}{n}\:{times}\:\mathrm{1}=\mathrm{11}...\mathrm{1} \\ $$$${N}_{\mathrm{2}} ={n}\:{times}\:\mathrm{2}=\mathrm{22}...\mathrm{2} \\ $$$${N}_{\mathrm{3}} ={n}\:{times}\:\mathrm{3}=\mathrm{33}...\mathrm{3} \\ $$$$ \\ $$$${N}_{\mathrm{1}} =\frac{\overset{\mathrm{2}{n}\:{times}} {\mathrm{99}...\mathrm{9}}}{\mathrm{9}}=\frac{\mathrm{10}^{\mathrm{2}{n}} −\mathrm{1}}{\mathrm{9}} \\ $$$${N}_{\mathrm{2}} =\frac{\mathrm{2}×\overset{{n}\:{times}} {\mathrm{99}...\mathrm{9}}}{\mathrm{9}}=\frac{\mathrm{2}\left(\mathrm{10}^{{n}} −\mathrm{1}\right)}{\mathrm{9}} \\ $$$${N}_{\mathrm{3}} =\frac{\overset{{n}\:{times}} {\mathrm{99}...\mathrm{9}}}{\mathrm{3}}=\frac{\mathrm{10}^{{n}} −\mathrm{1}}{\mathrm{3}} \\ $$$$ \\ $$$${N}_{\mathrm{1}} −{N}_{\mathrm{2}} =\frac{\mathrm{10}^{\mathrm{2}{n}} −\mathrm{1}−\mathrm{2}\left(\mathrm{10}^{{n}} −\mathrm{1}\right)}{\mathrm{9}} \\ $$$$=\frac{\left(\mathrm{10}^{{n}} \right)^{\mathrm{2}} −\mathrm{2}×\mathrm{10}^{{n}} +\mathrm{1}}{\mathrm{9}}=\left(\frac{\mathrm{10}^{{n}} −\mathrm{1}}{\mathrm{3}}\right)^{\mathrm{2}} \\ $$$$\sqrt{{N}_{\mathrm{1}} −{N}_{\mathrm{2}} }=\frac{\mathrm{10}^{{n}} −\mathrm{1}}{\mathrm{3}}={N}_{\mathrm{3}} \\ $$$$\Rightarrow{proved}! \\ $$

Commented by Pratah last updated on 30/Jan/20

thanks

$$\mathrm{thanks} \\ $$

Answered by naka3546 last updated on 31/Jan/20

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