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Question-9460




Question Number 9460 by tawakalitu last updated on 09/Dec/16
Answered by sou1618 last updated on 09/Dec/16
(√(11−2))=(√(10+1−2))=(√9)=3  (√(1111−22))=(√(1000+100+10+1−20−2))=(√(1089))=33  ......  X=(√(1111......11_(2000digits) −22...2_(1000digits) ))  ∗1000_(4digits) =10^(4−1)   X^2 =(10^(1999) +10^(1998) +...+10^1 +10^0 )−2(10^(999) +10^(998) +...+10^1 +10^0 )      =Σ_(k=1) ^(2000) 10^(k−1) −2Σ_(k=1) ^(1000) 10^(k−1)       =((10^(2000) −1)/(10−1))−2((10^(1000) −1)/(10−1))      =(1/9)(10^(2000) −2×10^(1000) +1)      =(1/3^2 ){(10^(1000) )^2 −2(10^(1000) )+1}      =(1/3^2 )(10^(1000) −1)^2       ={(1/3)(1000...000_(1001digits) −1)}^2   X=(1/3)×999...999_(1000digits)       =333...333_(1000digits)       (√(11...1_(2n(digits)) −22...2_(n(digits)) ))=33...3_(n(digits))
$$\sqrt{\mathrm{11}−\mathrm{2}}=\sqrt{\mathrm{10}+\mathrm{1}−\mathrm{2}}=\sqrt{\mathrm{9}}=\mathrm{3} \\ $$$$\sqrt{\mathrm{1111}−\mathrm{22}}=\sqrt{\mathrm{1000}+\mathrm{100}+\mathrm{10}+\mathrm{1}−\mathrm{20}−\mathrm{2}}=\sqrt{\mathrm{1089}}=\mathrm{33} \\ $$$$…… \\ $$$${X}=\sqrt{\mathrm{1111}……\mathrm{11}_{\mathrm{2000}{digits}} −\mathrm{22}…\mathrm{2}_{\mathrm{1000}{digits}} } \\ $$$$\ast\mathrm{1000}_{\mathrm{4}{digits}} =\mathrm{10}^{\mathrm{4}−\mathrm{1}} \\ $$$${X}^{\mathrm{2}} =\left(\mathrm{10}^{\mathrm{1999}} +\mathrm{10}^{\mathrm{1998}} +…+\mathrm{10}^{\mathrm{1}} +\mathrm{10}^{\mathrm{0}} \right)−\mathrm{2}\left(\mathrm{10}^{\mathrm{999}} +\mathrm{10}^{\mathrm{998}} +…+\mathrm{10}^{\mathrm{1}} +\mathrm{10}^{\mathrm{0}} \right) \\ $$$$\:\:\:\:=\underset{{k}=\mathrm{1}} {\overset{\mathrm{2000}} {\sum}}\mathrm{10}^{{k}−\mathrm{1}} −\mathrm{2}\underset{{k}=\mathrm{1}} {\overset{\mathrm{1000}} {\sum}}\mathrm{10}^{{k}−\mathrm{1}} \\ $$$$\:\:\:\:=\frac{\mathrm{10}^{\mathrm{2000}} −\mathrm{1}}{\mathrm{10}−\mathrm{1}}−\mathrm{2}\frac{\mathrm{10}^{\mathrm{1000}} −\mathrm{1}}{\mathrm{10}−\mathrm{1}} \\ $$$$\:\:\:\:=\frac{\mathrm{1}}{\mathrm{9}}\left(\mathrm{10}^{\mathrm{2000}} −\mathrm{2}×\mathrm{10}^{\mathrm{1000}} +\mathrm{1}\right) \\ $$$$\:\:\:\:=\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}} }\left\{\left(\mathrm{10}^{\mathrm{1000}} \right)^{\mathrm{2}} −\mathrm{2}\left(\mathrm{10}^{\mathrm{1000}} \right)+\mathrm{1}\right\} \\ $$$$\:\:\:\:=\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}} }\left(\mathrm{10}^{\mathrm{1000}} −\mathrm{1}\right)^{\mathrm{2}} \\ $$$$\:\:\:\:=\left\{\frac{\mathrm{1}}{\mathrm{3}}\left(\mathrm{1000}…\mathrm{000}_{\mathrm{1001}{digits}} −\mathrm{1}\right)\right\}^{\mathrm{2}} \\ $$$${X}=\frac{\mathrm{1}}{\mathrm{3}}×\mathrm{999}…\mathrm{999}_{\mathrm{1000}{digits}} \\ $$$$\:\:\:\:=\mathrm{333}…\mathrm{333}_{\mathrm{1000}{digits}} \\ $$$$ \\ $$$$ \\ $$$$\sqrt{\mathrm{11}…\mathrm{1}_{\mathrm{2}{n}\left({digits}\right)} −\mathrm{22}…\mathrm{2}_{{n}\left({digits}\right)} }=\mathrm{33}…\mathrm{3}_{{n}\left({digits}\right)} \\ $$
Commented by tawakalitu last updated on 09/Dec/16
Thank you sir. God bless you.
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{sir}.\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you}. \\ $$
Commented by tawakalitu last updated on 10/Dec/16
i really appreciate.
$$\mathrm{i}\:\mathrm{really}\:\mathrm{appreciate}. \\ $$

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