Cho

C

- Hướng dẫn giải

Đáp án C.

Ta có $S_n=5(1+11+111+\dots \underset{\text{n sè 1}}{\underbrace{111\dots 1}})$ 

Tính $A_n=1+11+111+\dots +\underset{\text{n sè 1}}{\underbrace{111\dots 1}}$ 

Xét:

$\begin{array}{l}{u}_1=1\\ u_2=1+10\\ u_3=1+10+{10}^2\\ u_4=1+10+{10}^2+{10}^3\\ ⋮\\ u_n=1+10+{10}^2+{10}^3+\dots +{10}^{n-1}\end{array}$

$A_n=n.1+(n-1)10+(n-2){.10}^2$$+\dots +\left[n-(n-1)\right]{10}^{n-1}$

$\Rightarrow 10A_n=10n+{10}^2(n-1)+{10}^3(n-2)$$+\dots +\left[n-(n-1)\right]{10}^n$

$\begin{array}{l}\left(2\right)-\left(1\right)\Rightarrow 9A_n\\ =\left(-n+10+{10}^2+{10}^3+\dots +{10}^{n-1}+{10}^n\right)\\ =\left(-n+10.\frac{1-{10}^n}{1-10}\right)\end{array}$

$\begin{array}{l}\Rightarrow A_n=\frac{1}{9}.\frac{{10}^{n+1}-10}{9}-\frac{n}{9}\\ \Rightarrow S_{2018}=5.A_{2018}\\ =\frac{5}{9}\left(\frac{{10}^{2019}-10}{9}-\frac{2018}{8}\right)\end{array}$