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Câu hỏi: Giá trị của n thỏa mãn: $C_{2n+1}^{1}-2.2C_{2n+1}^{2}+{{3.2}^{2}}.C_{2n+1}^{3}-{{4.2}^{3}}.C_{2n+1}^{4}+...+\left( 2n+1 \right){{.2}^{2n}}.C_{2n+1}^{2n+1}=2021$ bằng
A. 1010
B. 1009
C. 2020
D. 2021
A. 1010
B. 1009
C. 2020
D. 2021
Xét $kC_{2n+1}^{k}=k.\dfrac{\left( 2n+1 \right)!}{k!\left( 2n+1-k \right)!}=\dfrac{\left( 2n+1 \right).\left( 2n \right)!}{\left( k-1 \right)!\left[ 2n-\left( k-1 \right) \right]!}=\left( 2n+1 \right).C_{2n}^{k-1}$ với $1\le k\le 2n+1$.
Ta có $VT=\left( 2n+1 \right)\left( C_{2n}^{0}-C_{2n}^{1}{{2}^{1}}+C_{2n}^{2}{{2}^{2}}-...+C_{2n}^{2n}{{2}^{2n}} \right)=\left( 2n+1 \right){{\left( 1-2 \right)}^{2n}}=2n+1$.
Do đó: $2n+1=2021\Leftrightarrow n=1010$.
Ta có $VT=\left( 2n+1 \right)\left( C_{2n}^{0}-C_{2n}^{1}{{2}^{1}}+C_{2n}^{2}{{2}^{2}}-...+C_{2n}^{2n}{{2}^{2n}} \right)=\left( 2n+1 \right){{\left( 1-2 \right)}^{2n}}=2n+1$.
Do đó: $2n+1=2021\Leftrightarrow n=1010$.
Đáp án A.