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A

- Hướng dẫn giải

Đáp án A

$\begin{array}{l}y=x+x^2+x^3+\mathrm{...}+x^{2018}=x\frac{1-x^{2018}}{1-x}\\ y\text{'}=1+2x+3x^2+\mathrm{...}+2018x^{2017}\\ =\frac{2018x^{2019}-2019x^{2018}+1}{{(1-x)}^2}\\ y\text{'}\left(2\right)=1+2.2+{3.2}^2+\mathrm{...}+{2018.2}^{2017}={2017.2}^{2018}+1\end{array}$