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Question Number 45346 by Tawa1 last updated on 12/Oct/18

Commented by Meritguide1234 last updated on 12/Oct/18

this is my old post 45204

$${this}\:{is}\:{my}\:{old}\:{post}\:\mathrm{45204} \\ $$

Commented by Tawa1 last updated on 12/Oct/18

What is the solution

$$\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{solution} \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 12/Oct/18

=Σ_(r=1) ^(99) ((√(10+(√r) ))/(√(10−(√r) ))) (√(10+(√(99)))) > (√(10+(√1))) >(√(10+(√1))) (√(10+(√(99)))) > (√(10+(√2))) >(√(10+(√1))) (√(10+(√(99)))) >(√(10+(√3) )) >(√(10+(√1))) ........... ........... adding them 99×(√(10+(√(99)))) >N_r >99×(√(11)) x=10+(√(99)) 2x=20+2×(√(11)) ×3 2x=((√(11)) +3)^2 x=(1/2)((√(11)) +3)^2 (√x) =((3+(√(11)))/(√2)) =((3/(√2))+(√((11)/2)) ) 99×((3/(√2))+(√((11)/2)) )>N_r >99(√(11)) it is logic based justification...microsoft Excell can give the answer of the problem (√(10−(√(99)))) <(√(10−(√1))) <(√(10−(√1))) (√(10−(√(99)))) <(√(10−(√2))) <(√(10−(√1))) (√(10−(√(99)))) <(√(10−(√3))) <(√(10−(√1))) ........ ........ add them 99×(√(10−(√(99)))) <D_r <99×3 99×((√((11)/2)) −(3/(√2)))<D_r <99×3

$$=\underset{{r}=\mathrm{1}} {\overset{\mathrm{99}} {\sum}}\frac{\sqrt{\mathrm{10}+\sqrt{{r}}\:}}{\sqrt{\mathrm{10}−\sqrt{{r}}\:}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\sqrt{\mathrm{10}+\sqrt{\mathrm{99}}}\:\:\:\:>\:\:\sqrt{\mathrm{10}+\sqrt{\mathrm{1}}}\:\:\:\:\:>\sqrt{\mathrm{10}+\sqrt{\mathrm{1}}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\sqrt{\mathrm{10}+\sqrt{\mathrm{99}}}\:\:\:\:>\:\:\sqrt{\mathrm{10}+\sqrt{\mathrm{2}}}\:\:\:\:>\sqrt{\mathrm{10}+\sqrt{\mathrm{1}}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\sqrt{\mathrm{10}+\sqrt{\mathrm{99}}}\:\:\:\:>\sqrt{\mathrm{10}+\sqrt{\mathrm{3}}\:}\:\:\:\:\:\:>\sqrt{\mathrm{10}+\sqrt{\mathrm{1}}} \\ $$$$\:\:\:........... \\ $$$$\:\:\:\:\:...........\:\:\:{adding}\:{them} \\ $$$$\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\mathrm{99}×\sqrt{\mathrm{10}+\sqrt{\mathrm{99}}}\:\:\:>{N}_{{r}} \:\:\:>\mathrm{99}×\sqrt{\mathrm{11}}\: \\ $$$${x}=\mathrm{10}+\sqrt{\mathrm{99}}\: \\ $$$$\mathrm{2}{x}=\mathrm{20}+\mathrm{2}×\sqrt{\mathrm{11}}\:×\mathrm{3} \\ $$$$\mathrm{2}{x}=\left(\sqrt{\mathrm{11}}\:+\mathrm{3}\right)^{\mathrm{2}} \\ $$$${x}=\frac{\mathrm{1}}{\mathrm{2}}\left(\sqrt{\mathrm{11}}\:\:+\mathrm{3}\right)^{\mathrm{2}} \\ $$$$\sqrt{{x}}\:=\frac{\mathrm{3}+\sqrt{\mathrm{11}}}{\sqrt{\mathrm{2}}}\:=\left(\frac{\mathrm{3}}{\sqrt{\mathrm{2}}}+\sqrt{\frac{\mathrm{11}}{\mathrm{2}}}\:\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\mathrm{99}×\left(\frac{\mathrm{3}}{\sqrt{\mathrm{2}}}+\sqrt{\frac{\mathrm{11}}{\mathrm{2}}}\:\right)>{N}_{{r}} >\mathrm{99}\sqrt{\mathrm{11}}\: \\ $$$${it}\:{is}\:{logic}\:{based}\:{justification}...{microsoft}\: \\ $$$${Excell}\:{can}\:{give}\:{the}\:{answer}\:{of}\:{the}\:{problem} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\sqrt{\mathrm{10}−\sqrt{\mathrm{99}}}\:\:<\sqrt{\mathrm{10}−\sqrt{\mathrm{1}}}\:\:<\sqrt{\mathrm{10}−\sqrt{\mathrm{1}}}\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\sqrt{\mathrm{10}−\sqrt{\mathrm{99}}}\:\:<\sqrt{\mathrm{10}−\sqrt{\mathrm{2}}}\:\:<\sqrt{\mathrm{10}−\sqrt{\mathrm{1}}}\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\sqrt{\mathrm{10}−\sqrt{\mathrm{99}}}\:\:<\sqrt{\mathrm{10}−\sqrt{\mathrm{3}}}\:\:\:<\sqrt{\mathrm{10}−\sqrt{\mathrm{1}}}\: \\ $$$$\:\:\:\:\:\:\:\:\:........ \\ $$$$\:\:\:\:\:\:\:\:........\:\:\:\:{add}\:{them}\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{99}×\sqrt{\mathrm{10}−\sqrt{\mathrm{99}}}\:\:<{D}_{{r}} <\mathrm{99}×\mathrm{3} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{99}×\left(\sqrt{\frac{\mathrm{11}}{\mathrm{2}}}\:−\frac{\mathrm{3}}{\sqrt{\mathrm{2}}}\right)<{D}_{{r}} <\mathrm{99}×\mathrm{3} \\ $$

Commented by Tawa1 last updated on 12/Oct/18

God bless you sir. What is the final answer.

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}.\:\:\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{final}\:\mathrm{answer}. \\ $$

Commented by Meritguide1234 last updated on 12/Oct/18

ans is (√2)+1

$${ans}\:{is}\:\sqrt{\mathrm{2}}+\mathrm{1} \\ $$

Commented by Meritguide1234 last updated on 12/Oct/18

Commented by Tawa1 last updated on 12/Oct/18

God bless you sir

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$

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