"If people do not believe that mathematics is simple, it is only because they do not realize how complicated life is. ” J. von Neumann
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Nice exercise shared by Mikołaj Hajduk in their facebook profile
Remark:
If ${{a}_{n}}=\underbrace{111....11}_{n}$ then ${{a}_{n}}=\frac{{{10}^{n}}-1}{9}$ where $n\in \mathbb{N}$
Example : $\underbrace{11}_{2}=1\times 11=\frac{9}{9}\times 11=\frac{99}{9}=\frac{100-1}{9}=\frac{{{10}^{2}}-1}{9}=\frac{1}{9}\left( {{10}^{2}}-1 \right)$
Exercise:
Find the value of $w=\sqrt{\underbrace{44....4}_{2n}+\underbrace{11....1}_{n+1}-\underbrace{66....6}_{n}}$
Solution: we have $\underbrace{44....4}_{2n}=4\left( \underbrace{11....1}_{2n} \right)=4\left( \frac{{{10}^{2n}}-1}{9} \right)=\frac{4}{9}\left( {{10}^{2n}}-1 \right)$ , $\underbrace{11...1}_{n+1}=\frac{{{10}^{n+1}}-1}{9}$
and $\underbrace{66...6}_{n}=6\left( \underbrace{11...1}_{n} \right)=\frac{6}{9}\left( {{10}^{n}}-1 \right)$
so $\underbrace{44...4}_{2n}+\underbrace{11...1}_{n+1}-\underbrace{66...6}_{n}=\frac{4}{9}\left( {{10}^{2n}}-1 \right)+\frac{1}{9}\left( {{10}^{n+1}}-1 \right)-\frac{6}{9}\left( {{10}^{n}}-1 \right)$
$=\frac{1}{9}\left[ 4\left( {{10}^{2n}}-1 \right)+\left( {{10}^{n+1}}-1 \right)-6\left( {{10}^{n}}-1 \right) \right]$
$=\frac{1}{9}\left[ 4\times {{10}^{2n}}-4+{{10}^{n+1}}-1-6\times {{10}^{n}}+6 \right]$
$=\frac{1}{9}\left( 4\times {{10}^{2n}}+{{10}^{n+1}}+6\times {{10}^{n}}+1 \right)$
$=\frac{1}{9}\left( 4\times {{10}^{2n}} \right)+\frac{{{10}^{n+1}}}{9}-\frac{6\times {{10}^{n}}}{9}+\frac{1}{9}$
$=\frac{4}{9}{{\left( {{10}^{n}} \right)}^{2}}+\frac{10}{9}\left( {{10}^{n}} \right)-\frac{6}{9}\left( {{10}^{n}} \right)+\frac{1}{9}$
$={{\left( \frac{2}{3}{{10}^{n}} \right)}^{2}}+\frac{4}{9}{{10}^{n}}+\frac{1}{9}={{\left( \frac{2}{3}{{10}^{n}}+\frac{1}{3} \right)}^{2}}$
Thus $w=\sqrt{\underbrace{44...4}_{2n}+\underbrace{11...1}_{n+1}-\underbrace{66...6}_{n}}=\sqrt{{{\left( \frac{2}{3}{{10}^{n}}+\frac{1}{3} \right)}^{2}}}=\frac{2}{3}{{10}^{n}}+\frac{1}{3}=\frac{1}{3}\left( 2\times {{10}^{n}}+1 \right)=\underbrace{66...6}_{n-1}7$
Where $\frac{2}{3}\left( {{10}^{n}} \right)=0.6666...67\times {{10}^{n}}=\underbrace{66...6}_{n}.6667$and $0.66667+0.33333=1$
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The idea of Solution credit to Mikołaj Hajduk
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