Question-45346

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

Commented by Meritguide1234 last updated on 12/Oct/18

t h i s i s m y o l d p o s t 45204

Commented by Tawa1 last updated on 12/Oct/18

What is the solution

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

= \underset{r = 1}{\sum^{99}} \frac{\sqrt{10 + \sqrt{r}}}{\sqrt{10 - \sqrt{r}}}

\sqrt{10 + \sqrt{99}} > \sqrt{10 + \sqrt{1}} > \sqrt{10 + \sqrt{1}}

\sqrt{10 + \sqrt{99}} > \sqrt{10 + \sqrt{2}} > \sqrt{10 + \sqrt{1}}

\sqrt{10 + \sqrt{99}} > \sqrt{10 + \sqrt{3}} > \sqrt{10 + \sqrt{1}}

\dots \dots \dots . .

\dots \dots \dots . . a d d i n g t h e m

99 \times \sqrt{10 + \sqrt{99}} > N_{r} > 99 \times \sqrt{11}

x = 10 + \sqrt{99}

2 x = 20 + 2 \times \sqrt{11} \times 3

2 x = {(\sqrt{11} + 3)}^{2}

x = \frac{1}{2} {(\sqrt{11} + 3)}^{2}

\sqrt{x} = \frac{3 + \sqrt{11}}{\sqrt{2}} = (\frac{3}{\sqrt{2}} + \sqrt{\frac{11}{2}})

99 \times (\frac{3}{\sqrt{2}} + \sqrt{\frac{11}{2}}) > N_{r} > 99 \sqrt{11}

i t i s l o g i c b a s e d j u s t i f i c a t i o n \dots m i c r o s o f t

E x c e l l c a n g i v e t h e a n s w e r o f t h e p r o b l e m

\sqrt{10 - \sqrt{99}} < \sqrt{10 - \sqrt{1}} < \sqrt{10 - \sqrt{1}}

\sqrt{10 - \sqrt{99}} < \sqrt{10 - \sqrt{2}} < \sqrt{10 - \sqrt{1}}

\sqrt{10 - \sqrt{99}} < \sqrt{10 - \sqrt{3}} < \sqrt{10 - \sqrt{1}}

\dots \dots . .

\dots \dots . . a d d t h e m

99 \times \sqrt{10 - \sqrt{99}} < D_{r} < 99 \times 3

99 \times (\sqrt{\frac{11}{2}} - \frac{3}{\sqrt{2}}) < D_{r} < 99 \times 3

Commented by Tawa1 last updated on 12/Oct/18

God bless you sir . What is the final answer .

Commented by Meritguide1234 last updated on 12/Oct/18

a n s i s \sqrt{2} + 1

Commented by Meritguide1234 last updated on 12/Oct/18

Commented by Tawa1 last updated on 12/Oct/18

God bless you sir

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