$ (666 \ldots \text{up to } n \text{ digits})^2 + (888 \ldots \text{up to } n \text{ digits}) $ is:
Show Hint
- \( \frac{9}{4}(10^n - 1) \)
- \( \frac{9}{4}(10^n - 1)^2 \)
- \( \frac{4}{9}(10^{2n} + 1) \)
- \( \frac{4}{9}(10^{2n} - 1) \)
The Correct Option is D
Solution and Explanation
The given expression is \( (666 \ldots \text{up to } n \text{ digits})^2 + (888 \ldots \text{up to } n \text{ digits}) \).
We know that \( 666 \ldots \) up to \( n \) digits is represented as \( \frac{2}{3}(10^n - 1) \) and \( 888 \ldots \) up to \( n \) digits is represented as \( \frac{8}{9}(10^n - 1) \). Now, square \( 666 \ldots \): \[ \left( \frac{2}{3}(10^n - 1) \right)^2 = \frac{4}{9}(10^{2n} - 2 \times 10^n + 1). \] And for \( 888 \ldots \), we have: \[ \frac{8}{9}(10^n - 1). \] Adding both, the final result is: \[ \frac{4}{9}(10^{2n} - 1). \]
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