Question

$ (666 \ldots \text{up to } n \text{ digits})^2 + (888 \ldots \text{up to } n \text{ digits}) $ is:

• A. \( \frac{9}{4}(10^n - 1) \)
• B. \( \frac{9}{4}(10^n - 1)^2 \)
• C. \( \frac{4}{9}(10^{2n} + 1) \)
• D. \( \frac{4}{9}(10^{2n} - 1) \)

Hint:

For numbers like \( 666 \ldots \), use formulas to represent repeating digits, then apply basic algebra to manipulate and solve.

Answer 1

The Correct Option is D

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: \[ \frac{4}{9}(10^{2n} - 1). \]