Question

Find the sum to $n$ term of the series $6+66+666+....$

Answer 1

$6+66+666+...=6(1+11+111+...)=\frac{6}{9}(9+99+999+...)=\frac{6}{9}\left[\right(10-1)+({10}^2-1)+({10}^3-1)+...]$

$=\frac{6}{9}[10+{10}^2+{10}^3+.{.10}^n-(1+1+1+1+....x\left)\right]$

$=\frac{6}{9}\left[10\right(\frac{{10}^n-1}{10-1})-n]=\frac{6}{9}\left[\frac{10}{9}\right({10}^n-1)-n]$

$=\frac{6}{81}\frac{{10}^{n+1}-10-9n}{9}=\frac{6}{81}[{10}^{n+1}-9n-10]$