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

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(D) Let $n = 100$. The expression is $\sqrt{\sum_{k=0}^{2n-1} 10^k - 2 \sum_{k=0}^{n-1} 10^k}$.Using the sum of a geometric progression,$\sum_{k=0}^{m

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

Vedclass · Answer

(D) Let $n = 100$. The expression is $\sqrt{\sum_{k=0}^{2n-1} 10^k - 2 \sum_{k=0}^{n-1} 10^k}$.Using the sum of a geometric progression,$\sum_{k=0}^{m-1} 10^k = \frac{10^m - 1}{9}$.So,the expression becomes $\sqrt{\frac{10^{2n}-1}{9} - 2 \left( \frac{10^n-1}{9} ight)}$.$= \sqrt{\frac{10^{2n} - 1 - 2 \cdot 10^n + 2}{9}} = \sqrt{\frac{10^{2n} - 2 \cdot 10^n + 1}{9}}$.$= \sqrt{\left( \frac{10^n - 1}{3} ight)^2} = \frac{10^n - 1}{3}$.Since $\frac{10^n - 1}{9} = \underbrace{11...1}_{n 	ext{ digits}}$,then $\frac{10^n - 1}{3} = 3 	imes \underbrace{11...1}_{n 	ext{ digits}} = \underbrace{33...3}_{n 	ext{ digits}}$.For $n=100$,the result is $\underbrace{33...3}_{100 	ext{ digits}}$.

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

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