Phương pháp giải:

Sử dụng công thức: \(C_n^k + C_n^{k + 1} = C_{n + 1}^{k + 1};\,\,C_n^k = C_n^{n - k}\).

Lời giải chi tiết:

Từ \(C_n^k + C_n^{k + 1} = C_{n + 1}^{k + 1} \Rightarrow C_n^{k + 1} = C_{n + 1}^{k + 1} - C_n^k\)

\(\begin{array}{l}S = \sum\limits_{k = 0}^{1983} {C_{2017 + k}^k} \\ = \,\,\,\,\,C_{2017}^0 + \,\,\,\,\,\,\,\,\,\,\,C_{2018}^1\,\,\,\,\,\,\,\,\, + \,\,\,\,\,\,\,\,\,\,\,\,C_{2019}^2\,\,\,\,\,\,\,\,\, + ... + \,\,\,\,\,\,\,\,\,\,\,C_{4000}^{1983}\\ = \left( {C_{2018}^0} \right) + \left( {C_{2019}^1 - C_{2018}^0} \right) + \left( {C_{2020}^2 - C_{2019}^1} \right) + ... + \left( {C_{4001}^{1983} - C_{4000}^{1982}} \right)\\ = C_{4001}^{1983} = C_{4001}^{2018}\end{array}\)

Chọn: B