Explanation

(b)
$\begin{matrix}
(666\ldots 6) \\
n\ digits
\end{matrix} = 6 + 6 \times 10 + 6 \times 10^{2} + ... + 6 \times 10^{n - 1}$
$= 6(1 + 10 + 10^{2} + ... + 10^{n - 1})$
$= \frac{6}{9}\left( 10^{n} - 1 \right) = \frac{2}{3}(10^{n} - 1)$
Similarly, $\begin{matrix}
(888\ldots 8) \\
n\ digits
\end{matrix} = \frac{8}{9}(10^{n} - 1)$
Hence, required sum
$= \frac{4}{9}\left( 10^{n} - 1 \right)^{2} + \frac{8}{9}\left( 10^{n} - 1 \right)$
$= \frac{4}{9}(10^{2n} - {2.10}^{n} + 1 + {2.10}^{n} - 2)$
$= \frac{4}{9}(10^{2n} - 1)$