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Câu hỏi: Cho $n\in \mathbb{N}$ thỏa mãn $C_{n}^{n-1}+C_{n}^{n-2}=78.$ Tìm hệ số của ${{x}^{5}}$ trong khai triển ${{\left( 2x-1 \right)}^{n}}$.
A. 25344.
B. 101376.
C. $-101376.$
D. $-25344.$
A. 25344.
B. 101376.
C. $-101376.$
D. $-25344.$
Ta có
$C_{n}^{n-1}+C_{n}^{n-2}=78\Rightarrow \dfrac{n!}{\left( n-1 \right)!}+\dfrac{n!}{2!.\left( n-2 \right)!}=78\Rightarrow n+\dfrac{1}{2}\left( n-1 \right)n=78\Rightarrow n=12$
$\Rightarrow {{\left( 2x-1 \right)}^{12}}=\underset{k=0}{\overset{12}{\mathop \sum }} C_{12}^{k}{{\left( 2x \right)}^{12-k}}{{\left( -1 \right)}^{k}}=\underset{k=0}{\overset{12}{\mathop \sum }} C_{12}^{k}{{.2}^{12-k}}.{{\left( -1 \right)}^{k}}{{x}^{12-k}}$
$\Rightarrow 12-k=5\Rightarrow k=7\Rightarrow $ hệ số cần tìm là $C_{12}^{7}{{.2}^{5}}.\left( -1 \right)=-25344.$
$C_{n}^{n-1}+C_{n}^{n-2}=78\Rightarrow \dfrac{n!}{\left( n-1 \right)!}+\dfrac{n!}{2!.\left( n-2 \right)!}=78\Rightarrow n+\dfrac{1}{2}\left( n-1 \right)n=78\Rightarrow n=12$
$\Rightarrow {{\left( 2x-1 \right)}^{12}}=\underset{k=0}{\overset{12}{\mathop \sum }} C_{12}^{k}{{\left( 2x \right)}^{12-k}}{{\left( -1 \right)}^{k}}=\underset{k=0}{\overset{12}{\mathop \sum }} C_{12}^{k}{{.2}^{12-k}}.{{\left( -1 \right)}^{k}}{{x}^{12-k}}$
$\Rightarrow 12-k=5\Rightarrow k=7\Rightarrow $ hệ số cần tìm là $C_{12}^{7}{{.2}^{5}}.\left( -1 \right)=-25344.$
Đáp án D.