1708Yes Aditiya Great Solution
Thanks hsbhatt Sir
\hspace{-16}$If $\bf{x=\frac{\sqrt{10+\sqrt{1}}+\sqrt{10+\sqrt{2}}+\sqrt{10+\sqrt{3}}+.......+\sqrt{10+\sqrt{99}}}{\sqrt{10-\sqrt{1}}+\sqrt{10-\sqrt{2}}+\sqrt{10-\sqrt{3}}+.......+\sqrt{10-\sqrt{99}}}}$\\\\\\ Then value of $\bf{x}$ is
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3 Answers
341\sqrt{10+\sqrt 1} = \frac{1}{\sqrt 2} \left(\sqrt{10+\sqrt{99}} + \sqrt{10-\sqrt{99}}\right)
and
\sqrt{10-\sqrt 1} = \frac{1}{\sqrt 2} \left(\sqrt{10+\sqrt{99}} - \sqrt{10-\sqrt{99}}\right)
Thus the given expression can be written as
\frac{a+b}{a-b}
where
a = \sqrt{10+\sqrt 1} +\sqrt{10+\sqrt 2}+...+\sqrt{10+\sqrt {99}} and
b = \sqrt{10-\sqrt 1} +\sqrt{10-\sqrt 2}+...+\sqrt{10-\sqrt {99}}
so that we have
\frac{a}{b} = \frac{a+b}{a-b} \Rightarrow a^2 - 2ab -b^2 =0 \Rightarrow \frac{a}{b} = \boxed{\sqrt 2+1}