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Câu hỏi: Trong khai triển ${{\left( 1-3x \right)}^{n}}={{a}_{0}}+{{a}_{1}}x+{{a}_{2}}{{x}^{2}}+...+{{a}_{n}}{{x}^{n}}$. Tìm ${{a}_{2}}$ biết
${{a}_{0}}-{{a}_{1}}+{{a}_{2}}-{{a}_{3}}+...+{{\left( -1 \right)}^{n}}{{a}_{n}}={{2}^{2018}}$
A. ${{a}_{2}}=508536$
B. ${{a}_{2}}=9$
C. ${{a}_{2}}=4576824$
D. ${{a}_{2}}=18316377$
${{a}_{0}}-{{a}_{1}}+{{a}_{2}}-{{a}_{3}}+...+{{\left( -1 \right)}^{n}}{{a}_{n}}={{2}^{2018}}$
A. ${{a}_{2}}=508536$
B. ${{a}_{2}}=9$
C. ${{a}_{2}}=4576824$
D. ${{a}_{2}}=18316377$
Ta có: ${{\left( 1-3x \right)}^{n}}={{a}_{0}}+{{a}_{1}}x+{{a}_{2}}{{x}^{2}}+...+{{a}_{n}}{{x}^{n}}$
Thay $x=-1$ ta có: ${{4}^{n}}={{a}_{0}}-{{a}_{1}}+{{a}_{2}}...+{{\left( -1 \right)}^{n}}{{a}_{n}}={{2}^{2018}}={{4}^{1009}}\Rightarrow n=1009$
Xét khai triển ${{\left( 1-3x \right)}^{1009}}$ suy ra ${{a}_{2}}=C_{1009}^{2}.{{\left( 1 \right)}^{1007}}.{{\left( -3 \right)}^{2}}=4576824$.
Thay $x=-1$ ta có: ${{4}^{n}}={{a}_{0}}-{{a}_{1}}+{{a}_{2}}...+{{\left( -1 \right)}^{n}}{{a}_{n}}={{2}^{2018}}={{4}^{1009}}\Rightarrow n=1009$
Xét khai triển ${{\left( 1-3x \right)}^{1009}}$ suy ra ${{a}_{2}}=C_{1009}^{2}.{{\left( 1 \right)}^{1007}}.{{\left( -3 \right)}^{2}}=4576824$.
Đáp án C.