Google
Câu hỏi: Cho khai triển ${{\left( 3+x \right)}^{n}}={{a}_{0}}+{{a}_{1}}x+{{a}_{2}}{{x}^{2}}+...+{{a}_{n}}{{x}^{n}}$. Biết rằng ${{a}_{0}}-{{a}_{1}}+{{a}_{2}}-...+{{\left( -1 \right)}^{n}}{{a}_{n}}=4096$. Tìm ${{a}_{7}}$.
A. 192456.
B. 792.
C. 673596.
D. 1732104.
Từ khai triển ${{\left( 3+x \right)}^{n}}={{a}_{0}}+{{a}_{1}}x+{{a}_{2}}{{x}^{2}}+...+{{a}_{n}}{{x}^{n}}$ cho $x=-1$ ta có
${{\left( 3+\left( -1 \right) \right)}^{n}}={{a}_{0}}-{{a}_{1}}+{{a}_{2}}-...+{{\left( -1 \right)}^{n}}{{a}_{n}}=4096$ $\Leftrightarrow {{2}^{n}}=4096\Leftrightarrow n=12$
Ta có ${{\left( 3+x \right)}^{12}}=\sum\limits_{k=0}^{12}{C_{12}^{k}{{3}^{12-k}}{{\left( x \right)}^{k}}}$
Suy ra ${{a}_{7}}=C_{12}^{7}{{3}^{5}}=192456$.
A. 192456.
B. 792.
C. 673596.
D. 1732104.
Từ khai triển ${{\left( 3+x \right)}^{n}}={{a}_{0}}+{{a}_{1}}x+{{a}_{2}}{{x}^{2}}+...+{{a}_{n}}{{x}^{n}}$ cho $x=-1$ ta có
${{\left( 3+\left( -1 \right) \right)}^{n}}={{a}_{0}}-{{a}_{1}}+{{a}_{2}}-...+{{\left( -1 \right)}^{n}}{{a}_{n}}=4096$ $\Leftrightarrow {{2}^{n}}=4096\Leftrightarrow n=12$
Ta có ${{\left( 3+x \right)}^{12}}=\sum\limits_{k=0}^{12}{C_{12}^{k}{{3}^{12-k}}{{\left( x \right)}^{k}}}$
Suy ra ${{a}_{7}}=C_{12}^{7}{{3}^{5}}=192456$.
Đáp án A.