Question

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

Answer 1

Note that the given series is not a geometric series.
We need to find $S_n=6+66+666+...\text{to}n$ terms
$S_n=6(1+11+111+....\text{to}n$ terms )
$=\frac{6}{9}(9+99+999+....\text{to}n$ terms) (Multiply and divide by $9$)
$=\frac{2}{3}\left[\right(10-1)+(100-1)+(1000-1)+....ton\text{terms}]$
$=\frac{2}{3}\left[\right(10+{10}^2+{10}^3+....n\text{terms})-n]$
Thus, $S_n=\frac{2}{3}[\frac{10({10}^n-1)}{9}-n]$