Câu hỏi 4

\(f(x)=1+x+{{x}^{2}}+{{x}^{3}}+...+{{x}^{n}}=\frac{{{x}^{n+1}}-1}{x-1},\,\,\,(x\ne 1)\)

\(\begin{align}  f'(x)=1+2x+3{{x}^{2}}+...+n{{x}^{n-1}}=\frac{(n+1){{x}^{n}}(x-1)-({{x}^{n+1}}-1)}{{{\left( x-1 \right)}^{2}}}\\=\frac{(n+1){{x}^{n+1}}-(n+1){{x}^{n}}-{{x}^{n+1}}+1}{{{\left( x-1 \right)}^{2}}}=\frac{n{{x}^{n+1}}-(n+1){{x}^{n}}+1}{{{\left( x-1 \right)}^{2}}} \\ \end{align}\)

Cho \(x=2,\,\,n=2018\), ta có: \(S=1+2.2+{{3.2}^{2}}+{{4.2}^{3}}+...+{{2018.2}^{2017}}=\frac{{{2018.2}^{2019}}-{{2019.2}^{2018}}+1}{{{\left( 2-1 \right)}^{2}}}\\={{2}^{2018}}\left( 2018.2-2019 \right)+1={{2017.2}^{2018}}+1\)

Chọn: A