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Câu hỏi: Số phức $z=a+bi$ biết $z=1+i+{{i}^{2}}+2{{i}^{3}}+3{{i}^{4}}+...+2017{{i}^{2018}}$. Giá trị của $a+b$ là
A. 0
B. $-$ 2020
C. 3
D. 2018
A. 0
B. $-$ 2020
C. 3
D. 2018
Ta có $1+x+{{x}^{2}}+...+{{x}^{2017}}=\dfrac{{{x}^{2017}}-1}{x-1}$
$\begin{aligned}
& \Rightarrow 1+2x+...+2017{{x}^{2016}}=\dfrac{2017{{x}^{2016}}\left( x-1 \right)-{{x}^{2017}}+1}{{{\left( x-1 \right)}^{2}}} \\
& \Rightarrow z=1+i+{{i}^{2}}+2{{i}^{3}}+3{{i}^{4}}+...+2017{{i}^{2018}} \\
& =1+i+{{i}^{2}}\left( 1+2i+3{{i}^{2}}+...+2017{{i}^{2016}} \right) \\
& =1+i-\dfrac{2017{{i}^{2016}}\left( i-1 \right)-{{i}^{2017}}+1}{{{\left( i-1 \right)}^{2}}} \\
& =1+i+1008+1008i=1009+1009i \\
\end{aligned}$
$\begin{aligned}
& \Rightarrow 1+2x+...+2017{{x}^{2016}}=\dfrac{2017{{x}^{2016}}\left( x-1 \right)-{{x}^{2017}}+1}{{{\left( x-1 \right)}^{2}}} \\
& \Rightarrow z=1+i+{{i}^{2}}+2{{i}^{3}}+3{{i}^{4}}+...+2017{{i}^{2018}} \\
& =1+i+{{i}^{2}}\left( 1+2i+3{{i}^{2}}+...+2017{{i}^{2016}} \right) \\
& =1+i-\dfrac{2017{{i}^{2016}}\left( i-1 \right)-{{i}^{2017}}+1}{{{\left( i-1 \right)}^{2}}} \\
& =1+i+1008+1008i=1009+1009i \\
\end{aligned}$
Đáp án D.