The sum of n terms of the series $0.6+0.66+0.666+\cdots$ is

Sum of n terms of a G.P is $a.\frac{1-r^n}{1-r}$

Given series is $S_n=0.6+0.66+0.666+\cdots$

Taking 6 common we get $S_n=6(0.1+0.11+0.111+\cdots )$

Multiplying and dividing by 9 we get

$\begin{array}{l}{S}_n=\frac{6}{9}(0.9+0.99+0.999+\cdots )\\ =\frac{2}{3}[(1-0.1)+(1-0.01)+(1-0.001)+\cdots ]\\ =\frac{2}{3}[(1+1+1+\cdots )-(0.1+0.01+0.001+\cdots )]\end{array}$

But $1+1+1+\cdots =n$ and $0.1+0.01+0.001+\cdots$ is a G.P with 1^st term $=a=0.1$

And common ratio $=r=0.1$

We know that sum of n terms of a G.P is $a.\frac{1-r^n}{1-r}$

$\begin{array}{l}\therefore 0.1+0.01+0.001+\cdots =0.1\times \frac{1-{\left(0.1\right)}^n}{1-0.1}\\ 1-{\left(0.1\right)}^n=1-\frac{1}{{10}^n}=\frac{{10}^n-1}{{10}^n}\\ 1-0.1=1-\frac{1}{10}=\frac{9}{10}\\ \therefore 0.1+0.01+0.001+\cdots =\frac{1}{10}\times \frac{{10}^n-1}{{10}^n}\times \frac{10}{9}=\frac{{10}^n-1}{{9.10}^n}\end{array}$

Substituting the values in $S_n$ we get

$S_n=\frac{2}{3}\left[n-\frac{{10}^n-1}{{9.10}^n}\right]$

$=\frac{2n}{3}-\frac{2({10}^n-1)}{{27.10}^n}$