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Câu hỏi: Tính tổng $S=C_{n}^{0}-3C_{n}^{1}+{{3}^{2}}C_{n}^{2}-{{3}^{3}}C_{n}^{3}+...+{{\left( -1 \right)}^{n}}{{.3}^{n}}.C_{n}^{n}$
A. $-{{2}^{n}}$.
B. ${{\left( -2 \right)}^{n}}$.
C. ${{4}^{n}}$.
D. ${{2}^{n}}$.
A. $-{{2}^{n}}$.
B. ${{\left( -2 \right)}^{n}}$.
C. ${{4}^{n}}$.
D. ${{2}^{n}}$.
+Ta có $\forall n\in {{\mathbb{N}}^{*}}:{{\left(1+x \right)}^{n}}=C_{n}^{0}+xC_{n}^{1}+{{x}^{2}}C_{n}^{2}+{{x}^{3}}C_{n}^{3}+...+{{x}^{n}}C_{n}^{n}$.
Thay $x=-3$ vào hai vế ta được.
${{\left(1-3 \right)}^{n}}=C_{n}^{0}-3C_{n}^{1}+{{3}^{2}}C_{n}^{2}-{{3}^{3}}C_{n}^{3}+...+{{\left(-1 \right)}^{n}}{{. 3}^{n}}. C_{n}^{n}$
$\Leftrightarrow C_{n}^{0}-3C_{n}^{1}+{{3}^{2}}C_{n}^{2}-{{3}^{3}}C_{n}^{3}+...+{{\left(-1 \right)}^{n}}{{. 3}^{n}}. C_{n}^{n}={{\left(-2 \right)}^{n}}$.
Vậy tổng $S={{\left(-2 \right)}^{n}}$.
Thay $x=-3$ vào hai vế ta được.
${{\left(1-3 \right)}^{n}}=C_{n}^{0}-3C_{n}^{1}+{{3}^{2}}C_{n}^{2}-{{3}^{3}}C_{n}^{3}+...+{{\left(-1 \right)}^{n}}{{. 3}^{n}}. C_{n}^{n}$
$\Leftrightarrow C_{n}^{0}-3C_{n}^{1}+{{3}^{2}}C_{n}^{2}-{{3}^{3}}C_{n}^{3}+...+{{\left(-1 \right)}^{n}}{{. 3}^{n}}. C_{n}^{n}={{\left(-2 \right)}^{n}}$.
Vậy tổng $S={{\left(-2 \right)}^{n}}$.
Đáp án B.