Sum of n terms of the series 6 + 66 + 666 ... is

$$\begin{gathered} 6+66+666+.....=6(1+11+111+.....) \\ =\frac{6}{9}(9+99+999 .....) \\ =\frac{6}{9}\left(10-1+{10}^2-1+{10}^3-1.....\mathrm{n}+\mathrm{erms}\right) \\ =\frac{6}{9}\left(10\left(\frac{{10}^{\mathrm{n}}-1}{10-1}\right)-\mathrm{n}\right) \end{gathered}$$