Question

The sum of the first n terms of the series 6 + 66 + 666 + ….. is

• A.
• B.
• C.
• D.

Answer 1

The correct option is D

$6+66+666+....=\frac{2}{3}(9+99+999+...)=\frac{2}{3}\left[\right({10}^2-1)+({10}^3-1)+....]=\frac{2}{3}[10+{10}^2+{10}^3+....{.10}^n-n]=\frac{2}{3}[10\frac{({10}^n-1)}{(10-1)}-n]=\frac{2}{27}[{10}^{n+1}-10-9n]$