Tính S= 1+11+111+...+1111( với n số 1)

* Hướng dẫn giải

Đáp án D

Ta có $9S=9+99+999+\mathrm{...}+\underset{n\text{\hspace{0.17em}\hspace{0.17em}so\hspace{0.17em}9}}{\underbrace{\mathrm{99...99}}}$$=\left(10-1\right)+\left({10}^2-1\right)+$$\left({10}^3-1\right)+\mathrm{...}+\left({10}^n-1\right)$

$=\left(10+{10}^2+{10}^3+\mathrm{...}+{10}^n\right)-n$$=\frac{10\left(1-{10}^n\right)}{1-10}-n$$=\frac{10\left({10}^n-1\right)}{9}-n$

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