Question

The sum $\underset{ndigits}{\underbrace{(\mathrm{666...6}{)}^2}}$ +$\underset{ndigits}{\underbrace{(\mathrm{888...8})}}$is equal to

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

Answer 1

Answer: (C)

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