Tính tổng: 

D

- Hướng dẫn giải

Đáp án D

Ta có:

$\frac{k}{n}{\left(C_n^k\right)}^2=\frac{k}{n}{\left(\frac{n!}{k!\left(n-k\right)!}\right)}^2=C_n^k.\frac{\left(n-1\right)!}{\left(k-1\right)!\left(n-k\right)!}=C_n^k.C_{n-1}^{k-1}$

Do đó: $C_{2018}^0.C_{2018}^1+C_{2018}^1.C_{2018}^2+\mathrm{...}+C_{2018}^{2017}.C_{2018}^{2018}$

Xét khai triển: ${\left(1+x\right)}^{2018}.\left(x+1\right)={\left(1+x\right)}^{4036}$

Hệ số chứa $x^{2017}$ trong khai triển ${\left(1+x\right)}^{2018}.\left(x+1\right)$ là:

$C_{2018}^0.C_{2018}^1+C_{2018}^1.C_{2018}^2+\mathrm{...}+C_{2018}^{2017}.C_{2018}^{2018}=S$

Hệ số chứa $x^{2017}$ trong khai triển ${\left(1+x\right)}^{4036}$ là:

$C_{4036}^{2017}=\frac{4036!}{2017!.2019!}=\frac{4036!}{2018!.2018!}.\frac{2018}{2019}=\frac{2018}{2019}{C}_{4036}^{2018}$

Vậy $S=\frac{2018}{2019}{C}_{4036}^{2018}$