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A

- Hướng dẫn giải

Đáp án A

${(1+x)}^{2017}=C_{2017}^0+C_{2017}^{1}x+\mathrm{...}+C_{2017}^{2017}{x}^{2017}$

Lấy đạo hàm hai vế ta được:

$\begin{array}{l}2017{\left(1+x\right)}^{2016} =C_{2017}^1+2C_{2017}^{2}x+\mathrm{...}+2017C_{2017}^{2017}{x}^{2016}\\ {(1+x)}^{2016}=\frac{1}{2017}(C_{2017}^1+2C_{2017}^{2}x+\mathrm{...}+2017C_{2017}^{2017}{x}^{2016})\\ x=3\Rightarrow 4^{2016}=1+\frac{1}{2017}(2.3C_{2017}^2+\mathrm{...}+{2017.3}^{2016}{C}_{2017}^{2017})\\ \Rightarrow \frac{1}{2017}(2.3C_{2017}^2+\mathrm{...}+{2017.3}^{2016}{C}_{2017}^{2017})=4^{2016}-1\end{array}$