Phương pháp giải:

Áp dụng công thức: \(A_n^k = \frac{{n!}}{{(n - k)!}}\)

Chú ý \(n!.n = (n + 1 - 1)n! = (n + 1)! - n!\)

Từ đó rút gọn biểu thức đưa về dang đơn giản.

Lời giải chi tiết:

\(\begin{array}{l}\;\;\;\;\;P = \frac{{2017}}{{A_{2017}^0}} + \frac{{2016}}{{A_{2017}^1}} + ...... + \frac{2}{{A_{2017}^{2015}}} + \frac{1}{{A_{2017}^{2016}}}\\ \Leftrightarrow P = \frac{{2017}}{{\frac{{2017!}}{{\left( {2017 - 0} \right)!}}}} + \frac{{2016}}{{\frac{{2017!}}{{\left( {2017 - 1} \right)!}}}} + .......... + \frac{2}{{\frac{{2017!}}{{\left( {2017 - 2015} \right)!}}}} + \frac{1}{{\frac{{2017!}}{{\left( {2017 - 2016} \right)!}}}}\\ \Leftrightarrow P = \frac{{2017.2017!}}{{2017!}} + \frac{{2016.2016!}}{{2017!}} + ... + \frac{{2.2!}}{{2017!}} + \frac{{1.1!}}{{2017!}}\\ \Leftrightarrow P = \frac{{2017.2017! + 2016.2016! + ... + 2.2! + 1.1!}}{{2017!}}\\ \Leftrightarrow P = \frac{{\left( {2018 - 1} \right)2017! + \left( {2017 - 1} \right)2016! + ... + \left( {3 - 1} \right)2! + \left( {2 - 1} \right)1!}}{{2017!}}\\ \Leftrightarrow P = \frac{{2018.2017! - 2017! + 2017.2016! - 2016! + ........ + 3.2! - 2! + 2! - 1!}}{{2017!}}\\ \Leftrightarrow P = \frac{{\left( {2018! - 2017!} \right) + \left( {2017! - 2016!} \right) + ... + \left( {3! - 2!} \right) + \left( {2! - 1!} \right)}}{{2017!}} = \frac{{2018! - 1!}}{{2017!}}\\ \Leftrightarrow P = 2018 - \frac{1}{{2017!}}.\end{array}\)

Chọn C