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 

The Gif file is done by

Mikołaj Hajduk