$\begin{array}{*{20}{c}} {{{\left( {666 \ldots 6} \right)}^2}}\\ {n\;{\rm{digits}}} \end{array} + \begin{array}{*{20}{c}} {\left( {888 \ldots 8} \right)}\\ {n\;{\rm{digits}}} \end{array}$ is equal to

$\frac{4}{9}\left( {{{10}^n} - 1} \right)$

$\frac{4}{9}{\left( {{{10}^n} - 1} \right)^2}$

JEE MATHEMATICS MCQ: SEQUENCE OF SERIES

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Application +4 | -1

\begin{array}{cccccccccccccccccccc} {{{\left( {666 \ldots 6} \right)}^2}}\\ {n\;\mathrm{digits}} \end{array} + \begin{array}{cccccccccccccccccccc} {\left( {888 \ldots 8} \right)}\\ {n\;\mathrm{digits}} \end{array}

is equal to

View answer and explanation

Correct option: B

\frac{4}{9}\left( {{{10}^{2n}} - 1} \right)

Option B (

\frac{4}{9}\left( {{{10}^{2n}} - 1} \right)

) is correct because, by converting the repeating digits into fractions and simplifying, the expression equals

\frac{4}{9}\left( {{{10}^{2n}} - 1} \right)