sqrt {underbrace {111........1}_{200,text{dig | vedclass

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(D) Let $n = 100$. The expression is $\sqrt {\frac{10^{2n}-1}{9} - 2\left( \frac{10^n-1}{9} ight)}$.$= \sqrt {\frac{10^{2n}-1 - 2 	imes 10^n + 2}{9

Vedclass · Accepted Answer

$\underbrace {33.......3}_{100\,	ext{digits}}$

Vedclass · Answer

(D) Let $n = 100$. The expression is $\sqrt {\frac{10^{2n}-1}{9} - 2\left( \frac{10^n-1}{9} ight)}$.$= \sqrt {\frac{10^{2n}-1 - 2 	imes 10^n + 2}{9}} = \sqrt {\frac{10^{2n} - 2 	imes 10^n + 1}{9}}$.$= \sqrt {\left( \frac{10^n-1}{3} ight)^2} = \frac{10^n-1}{3}$.Since $\frac{10^{100}-1}{9} = \underbrace{111...1}_{100\,	ext{digits}}$,then $\frac{10^{100}-1}{3} = 3 	imes \underbrace{111...1}_{100\,	ext{digits}} = \underbrace{333...3}_{100\,	ext{digits}}$.

$\sqrt {\underbrace {111........1}_{200\,\text{digits}} - \underbrace {222.......2}_{100\,\text{digits}}} $ equals :-

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