Đáp án:
A
Giải thích các bước giải:
Ta có \(2017!.(1+\frac{1}{1})^{1}(1+\frac{1}{2})^{2}...(1+\frac{1}{2017})^{2017}
= 2017!.2^1.(\frac{3}{2})^2.(\frac{4}{3})^3....(\frac{2017}{2016})^{2016}.(\frac{2018}{2017})^{2017}
= 2017!.\frac{1}{1}.\frac{1}{2}.\frac{1}{3}...\frac{1}{2016}.\frac{2018^{2017}}{2017}
= 2018^{2017} = a^{b}\Rightarrow (a;b)=(2018;2017)\)