Question

Find the sum of the first $n$ terms of the series $0.2+0.22+0.222+\dots$

Answer 1

Let $S=0.2+0.22+0.222+\dots$
$\therefore S_n=\frac{2}{9}[0.9+0.99+0.999+0.9999+\cdots +nterms]$
$=\frac{2}{9}[(1-0.1)+(1-0.01)+(1-0.001)+\cdots +nterms]$
$=\frac{2}{9}[n-(0.1+0.01+\cdots +nterms)]$
$=\frac{2n}{9}-\frac{2}{9}\frac{\left(0.1\right)[1-{\left(0.1\right)}^n]}{[1-0.1]}$
$=\frac{2}{9}n-\frac{2}{9}\frac{\left(0.1\right)}{0.9}[1-{\left(0.1\right)}^n]$
$=\frac{2}{9}n-\frac{2}{81}+\frac{2}{81}{\left(0.1\right)}^n$