Question

Prove the equality:
$\underset{n\text{ digits}}{\mathop{{{\left( 666...6 \right)}^{2}}}}\,+\underset{n\text{ digits}}{\mathop{888...8}}\,=\underset{2n\text{ digits}}{\mathop{444...4}}\,$

Answer 1

Hint: Take ${{S}_{1}}=666...6,{{S}_{2}}=888...8$. Expand them to form a geometric progression. Then find the value of ${{S}_{1}}^{2}+{{S}_{2}}$.

“Complete step-by-step answer:”
$\begin{aligned} & \text{Take }{{S}_{1}}=666...6 \\\ & \text{Given }{{S}_{1}}=666...\text{upto n-digits} \\\ & =6+6\times 10+6\times {{10}^{2}}+6\times {{10}^{3}}+...+6\times {{10}^{n-1}} \\\ & =6\left( 1+10+{{10}^{2}}+....+{{10}^{n-1}} \right) \\\ \end{aligned}$
Now $1+10+{{10}^{2}}+....+{{10}^{n-1}}$ is the expanded form of $\dfrac{{{10}^{n}}-1}{10-1}$
i.e., It is of the form $\dfrac{a\left( {{r}^{n}}-1 \right)}{r-1}$
Where a = first term, r = common ratio
Which is a geometric progression.
Here a = 1, r = 10.
$\therefore {{S}_{1}}=6\left[ 1\left( \dfrac{{{10}^{n}}-1}{10-1} \right) \right]=6\times \dfrac{\left( {{10}^{n}}-1 \right)}{9}=\dfrac{2}{3}\left( {{10}^{n}}-1 \right)$
Now ${{S}_{2}}=888...$ up to n-digits

& =8+8\times 10+8\times {{10}^{2}}+...+8\times {{10}^{n-1}} \\\ & =8\left( 1+10+{{10}^{2}}+...+{{10}^{n-1}} \right) \\\ & =8\times 1\left[ \dfrac{{{10}^{n}}-1}{10-1} \right]=\dfrac{8}{9}\left( {{10}^{n}}-1 \right) \\\ \end{aligned}$$ We need to find ${{\left( {{S}_{1}} \right)}^{2}}+{{S}_{2}}={{\left[ \dfrac{2}{3}\left( {{10}^{n}}-1 \right) \right]}^{2}}+\dfrac{8}{9}\left( {{10}^{n}}-1 \right)$ $\begin{aligned} & =\dfrac{4}{9}{{\left( {{10}^{n}}-1 \right)}^{2}}+\dfrac{8}{9}\left( {{10}^{n}}-1 \right) \\\ & =\dfrac{\left( {{10}^{n}}-1 \right)}{9}\left[ 4\left( {{10}^{n}}-1 \right)+8 \right] \\\ \end{aligned}$ Taking 4 common, $\begin{aligned} & =4\left( \dfrac{{{10}^{n}}-1}{10-1} \right)\left[ \left( {{10}^{n}}-1 \right)+2 \right] \\\ & =4\left[ \dfrac{{{10}^{n}}-1}{10-1} \right]\left[ {{10}^{n}}-1+2 \right] \\\ & =4\dfrac{\left( {{10}^{n}}-1 \right)\left( {{10}^{n}}+1 \right)}{10-1} \\\ & =4\dfrac{\left[ {{\left( {{10}^{n}} \right)}^{2}}-{{1}^{2}} \right]}{10-1} \\\ \end{aligned}$ We know ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$ $=\dfrac{4\left[ {{10}^{2n}}-1 \right]}{\left[ 10-1 \right]}$ $\therefore $444....4 up to 2n digits. Hence proved. Note: A geometric progression is a sequence in which each term is derived by multiplying/dividing the preceding term by a fixed number called common ratio. The general form of GP is $a,ar,a{{r}^{2}},a{{r}^{3}}$ and so on; a = first term r = common ratio