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Câu hỏi: Tìm số hạng không chứa $x$ trong khai triển ${{\left( x-\dfrac{2}{x} \right)}^{n}},n\in {{\mathbb{N}}^{*}}$ biết $C_{n}^{1}-2.2.C_{n}^{2}+{{3.2}^{2}}.C_{n}^{3}-{{4.2}^{3}}.C_{n}^{4}+{{5.2}^{4}}C_{n}^{5}+...+{{\left( -1 \right)}^{n}}.n{{.2}^{n-1}}C_{n}^{n}=-2022$
A. $-C_{2021}^{1009}{{2}^{1009}}$
B. $-C_{2018}^{1009}{{2}^{1009}}$
C. $C_{2020}^{1010}{{2}^{1010}}$
D. $-C_{2022}^{1011}{{2}^{1011}}$
A. $-C_{2021}^{1009}{{2}^{1009}}$
B. $-C_{2018}^{1009}{{2}^{1009}}$
C. $C_{2020}^{1010}{{2}^{1010}}$
D. $-C_{2022}^{1011}{{2}^{1011}}$
Xét khai triển:
${{\left( 1-x \right)}^{n}}=\sum\limits_{k=0}^{n}{C_{n}^{k}{{\left( -x \right)}^{k}}}$
$=C_{n}^{0}-C_{n}^{1}.x+C_{n}^{2}.{{x}^{2}}-C_{n}^{3}.{{x}^{3}}+...+{{\left( -1 \right)}^{k}}.{{x}^{k}}.C_{n}^{k}+...+C_{n}^{2}.{{\left( -x \right)}^{n}}$
Lấy đạo hàm cả hai vế ta được:
$-n{{\left( 1-x \right)}^{n-1}}=-C_{n}^{1}+2.C_{n}^{2}.x-3.{{x}^{2}}.C_{n}^{3}+...+{{\left( -1 \right)}^{k}}.k.{{x}^{k-1}}.C_{n}^{k}+...-C_{n}^{n}.n.{{\left( -x \right)}^{n-1}}$
$\Rightarrow n{{\left( 1-x \right)}^{n-1}}=C_{n}^{1}-2.x.C_{n}^{2}+3.{{x}^{2}}.C_{n}^{3}-...-{{\left( -1 \right)}^{k}}.k.{{x}^{k-1}}.C_{n}^{k}-...+C_{n}^{n}.n.{{\left( -x \right)}^{n-1}}$
Cho $x=2$ ta được
$n.{{\left( -1 \right)}^{n-1}}=C_{n}^{1}-2.2.C_{n}^{2}+{{3.2}^{2}}.C_{n}^{3}-{{4.2}^{3}}.C_{n}^{4}+{{5.2}^{4}}.C_{n}^{5}+...+{{\left( -1 \right)}^{n}}.n{{.2}^{n-1}}.C_{n}^{n}$
$\Leftrightarrow n.{{\left( -1 \right)}^{n-1}}=-2022\Leftrightarrow n=2022$
Xét khai triển: ${{\left( x-\dfrac{2}{x} \right)}^{2020}}=\sum\limits_{k=0}^{2022}{C_{2022}^{k}.{{x}^{2022-k}}.{{\left( \dfrac{-2}{x} \right)}^{k}}}$
$=\sum\limits_{k=0}^{2022}{C_{2022}^{k}.{{\left( -2 \right)}^{k}}.{{x}^{2022-2k}}}$
Số hạng không chứa $x$ ứng với: $2022-2k=0$
$\Leftrightarrow k=1011$
Vậy số hạng không chứa $x$ là: $-C_{2022}^{1011}{{.2}^{1011}}$
${{\left( 1-x \right)}^{n}}=\sum\limits_{k=0}^{n}{C_{n}^{k}{{\left( -x \right)}^{k}}}$
$=C_{n}^{0}-C_{n}^{1}.x+C_{n}^{2}.{{x}^{2}}-C_{n}^{3}.{{x}^{3}}+...+{{\left( -1 \right)}^{k}}.{{x}^{k}}.C_{n}^{k}+...+C_{n}^{2}.{{\left( -x \right)}^{n}}$
Lấy đạo hàm cả hai vế ta được:
$-n{{\left( 1-x \right)}^{n-1}}=-C_{n}^{1}+2.C_{n}^{2}.x-3.{{x}^{2}}.C_{n}^{3}+...+{{\left( -1 \right)}^{k}}.k.{{x}^{k-1}}.C_{n}^{k}+...-C_{n}^{n}.n.{{\left( -x \right)}^{n-1}}$
$\Rightarrow n{{\left( 1-x \right)}^{n-1}}=C_{n}^{1}-2.x.C_{n}^{2}+3.{{x}^{2}}.C_{n}^{3}-...-{{\left( -1 \right)}^{k}}.k.{{x}^{k-1}}.C_{n}^{k}-...+C_{n}^{n}.n.{{\left( -x \right)}^{n-1}}$
Cho $x=2$ ta được
$n.{{\left( -1 \right)}^{n-1}}=C_{n}^{1}-2.2.C_{n}^{2}+{{3.2}^{2}}.C_{n}^{3}-{{4.2}^{3}}.C_{n}^{4}+{{5.2}^{4}}.C_{n}^{5}+...+{{\left( -1 \right)}^{n}}.n{{.2}^{n-1}}.C_{n}^{n}$
$\Leftrightarrow n.{{\left( -1 \right)}^{n-1}}=-2022\Leftrightarrow n=2022$
Xét khai triển: ${{\left( x-\dfrac{2}{x} \right)}^{2020}}=\sum\limits_{k=0}^{2022}{C_{2022}^{k}.{{x}^{2022-k}}.{{\left( \dfrac{-2}{x} \right)}^{k}}}$
$=\sum\limits_{k=0}^{2022}{C_{2022}^{k}.{{\left( -2 \right)}^{k}}.{{x}^{2022-2k}}}$
Số hạng không chứa $x$ ứng với: $2022-2k=0$
$\Leftrightarrow k=1011$
Vậy số hạng không chứa $x$ là: $-C_{2022}^{1011}{{.2}^{1011}}$
Đáp án D.