(b)
Let $S = 6 + 66 + 666 + ...n$ terms
$= \frac{6}{9}(9 + 99 + 999 + ...n$ terms)
$= \frac{2}{3}\lbrack(10 - 1) + (100 - 1) + (1000 - 1) + ...n$ terms]
$= \frac{2}{3}\left\lbrack 10 \bullet \frac{{(10}^{n} - 1)}{9} - n \right\rbrack = \frac{2}{27}\lbrack 10^{n + 1} - 10 - 9n\rbrack$