Tổng

Câu hỏi: Tổng $S=1+11+111+\mathrm{...}+\underset{n\text{\hspace{0.17em}\hspace{0.17em}so\hspace{0.17em}1}}{\underbrace{\mathrm{11...111}}}$ là

A. $S=\frac{10}{81}\left({10}^{n-1}-1\right)-\frac{n}{9}.$

B. $S=\frac{10}{81}\left({10}^n-1\right)+\frac{n}{9}.$

C. $S=\frac{1}{81}\left({10}^n-1\right)-\frac{n}{9}.$

D. $S=\frac{10}{81}\left({10}^n-1\right)-\frac{n}{9}.$