Question

Prove that

(666...n digits)^2+(888...ndigits) = (444...2ndigits)​

Answer 1

$\large\underline{\sf{Solution-}}$

Consider LHS

$\rm \: {(666 \cdots \: n \: digits)}^{2} + (888 \cdots \: n \: digits)$

Let first evaluate both the terms individually.

So, Consider first term

$\rm \: {(666 \cdots \: n \: digits)}^{2}$

$\rm \: = {[6(111 \cdots \: n \: digits)]}^{2}$

$\rm \: = {\bigg[\dfrac{6}{9} (999 \cdots \: n \: digits)\bigg]}^{2}$

$\rm \: = {\bigg[\dfrac{2}{3} ( {10}^{n} - 1) \bigg]}^{2}$

$\rm \: = \dfrac{4}{9} {( {10}^{n} - 1)}^{2}$

$\red{\rm\implies \: \boxed{ \rm{ \:{(666 \cdots \: n \: digits)}^{2} = \dfrac{4}{9} {( {10}^{n} - 1)}^{2}} \: }}$

Now, Consider second term

$\rm \: {(888 \cdots \: n \: digits)}$

$\rm \: = 8(111 \cdots \: n \: digits)$

$\rm \: = \dfrac{8}{9} (999\cdots \: n \: digits)$

$\rm \: = \dfrac{8}{9} ( {10}^{n} - 1)$

So,

$\red{\rm\implies \:\boxed{ \rm{ \:(888 \cdots \: n \: digits) = \dfrac{8}{9} ( {10}^{n} - 1) \: }}}$

Now, Consider LHS

$\rm \: {(666 \cdots \: n \: digits)}^{2} + (888 \cdots \: n \: digits)$

On substituting the values of both the terms evaluated above, we get

$\rm \: = \: \dfrac{4}{9} {( {10}^{n} - 1)}^{2} + \dfrac{8}{9}( {10}^{n} - 1)$

$\rm \: = \: \dfrac{4}{9}( {10}^{n} - 1)\bigg( {( {10}^{n} - 1)} + 2\bigg)$

$\rm \: = \: \dfrac{4}{9}( {10}^{n} - 1)\bigg( {{10}^{n} - 1} + 2\bigg)$

$\rm \: = \: \dfrac{4}{9}( {10}^{n} - 1)\bigg( {{10}^{n} + 1}\bigg)$

$\rm \: = \: \dfrac{4}{9}( {10}^{2n} - 1)$

$\rm \: = \: \dfrac{4}{9}(999 \cdots \: 2n \: digits)$

$\rm \: = \: 4(111\cdots \: 2n \: digits)$

$\rm \: = \: (444\cdots \: 2n \: digits)$

Hence,

$\red{\rm\implies \:\rm \: {(666 \cdots \: n \: digits)}^{2} + (888 \cdots \: n \: digits)} \\ \\ \rm \: \: \: \red{ = \: (444 \cdots \: 2n \: digits)} \: \: \:$

$\rule{190pt}{2pt}$

Concept Used :-

$\rm \: 9 = 10 - 1$

$\rm \: 99 = 100 - 1 = {10}^{2} - 1$

$\rm \: 999 = 1000 - 1 = {10}^{3} - 1$

$\rm \: 9999 = 10000 - 1 = {10}^{4} - 1$

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$\rm \: (9999 \cdots \: n \: digits) = {10}^{n} - 1$

Answer 2

Answer:

$\boxed{\underline{\text { Solution : - }} \: \: \:}$

Step-by-step explanation:

$\begin{aligned} \text{L.H.S. } & \rm=(666 \ldots \text { upto } n \text { digits })^{2} \\ & \rm+888 \text {...upto } n \text { digits } \\ \\ =& \rm(6+60+600+\ldots \text { upto } n \text { terms })^{2} \\ & \rm+(8+80+800+\ldots \text { upto } n \text { terms }) \\ \\ =& \rm6^{2}\left\{\frac{10^{n}-1}{10-1}\right\}^{2}+8\left\{\frac{10^{n}-1}{10-1}\right\} \\ \\ =& \rm 36\left(\frac{10^{2 n}-2 \times 10^{n}+1}{81}\right)+\frac{8\left(10^{n}-1\right)}{9} \\ \\ =& \rm\frac{4\left(10^{2 n}-2 \times 10^{n}+1\right)+8 \times 10^{n}-8}{9} \\ \\ =& \rm \frac{4 \times 10^{2 n}-4}{9}=\frac{4}{9}\left(10^{2 n}-1\right)=\frac{4\left(10^{2 n}-1\right)}{10-1} \\ \\ =& \rm 4+40+400+\ldots \text { upto } 2 n \text { terms } \\ \\\color{orange} =& \color{orange} \rm 4444 \ldots \text {...upto } 2 n \text { digits }=\text { R.H.S. } \huge _{_{//} } \end{aligned}$