Question

Sum of n terms of the series $ 1 + 11 + 111 + ...$ is

• A. $\frac{10}{81}(10^n-1)-\frac{n}{81}$
• B. $\frac{10}{81}(10^n-1)-\frac{n}{9}$
• C. $\frac{10}{81}(10^n-1)+\frac{n}{9}$
• D. $\frac{10}{81}(10^n-1)+\frac{n}{81}$

Answer 1

Answer: (B)

$S_{n} = 1+11+111+.....n$ terms
$ = \frac{1}{9}[9+99+999...n $ terms]
$= \frac{1}{9} [\left(10-1\right)+\left(10^{2}-1\right)+\left(10^{3}-1\right)+...$)
$ = \frac{1}{9}\left[10 \frac{\left(10^{n}-1\right)}{10-1}-n\right]$
$= \frac{10}{81}\left[10^{n}-1\right]-\frac{n}{9}$