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Câu hỏi: Giá trị biểu thức $P=C_{2019}^{0}+2C_{2019}^{1}+3C_{2019}^{2}+...+2020C_{2019}^{2019}$ bằng
A. ${{2}^{2019}}$
B. ${{2019.2}^{2018}}$
C. ${{2020.2}^{2018}}$
D. ${{2021.2}^{2018}}$
A. ${{2}^{2019}}$
B. ${{2019.2}^{2018}}$
C. ${{2020.2}^{2018}}$
D. ${{2021.2}^{2018}}$
HD: Ta có: ${{\left( 1+x \right)}^{2019}}=C_{2019}^{0}+C_{2019}^{1}x+C_{2019}^{2}{{x}^{2}}+...+C_{2019}^{2019}{{x}^{2019}}$
Nhân cả 2 vế với x ta được: $x{{\left( 1+x \right)}^{2019}}=xC_{2019}^{0}+C_{2019}^{1}{{x}^{2}}+C_{2019}^{2}{{x}^{3}}+...+C_{2019}^{2019}{{x}^{2020}}$
Đạo hàm 2 vế ta có: ${{\left( 1+x \right)}^{2019}}+2019x{{\left( 1+x \right)}^{2018}}=C_{2019}^{0}+2xC_{2019}^{1}+...+2020{{x}^{2019}}C_{2019}^{2019}$
Thay $x=1\Rightarrow {{2}^{2019}}+{{2019.2}^{2018}}=P\Rightarrow P={{2021.2}^{2018}}.$
Nhân cả 2 vế với x ta được: $x{{\left( 1+x \right)}^{2019}}=xC_{2019}^{0}+C_{2019}^{1}{{x}^{2}}+C_{2019}^{2}{{x}^{3}}+...+C_{2019}^{2019}{{x}^{2020}}$
Đạo hàm 2 vế ta có: ${{\left( 1+x \right)}^{2019}}+2019x{{\left( 1+x \right)}^{2018}}=C_{2019}^{0}+2xC_{2019}^{1}+...+2020{{x}^{2019}}C_{2019}^{2019}$
Thay $x=1\Rightarrow {{2}^{2019}}+{{2019.2}^{2018}}=P\Rightarrow P={{2021.2}^{2018}}.$
Đáp án D.