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Câu hỏi: Cho dãy số $\left( {{u}_{n}} \right):\left\{ \begin{aligned}
& {{u}_{1}}=-3 \\
& {{u}_{n+1}}={{u}_{n}}+\dfrac{5}{2},\ n\ge 1 \\
\end{aligned} \right. $. Tính $ S={{u}_{20}}-{{u}_{6}}$.
A. $S=33.$
B. $S=\dfrac{69}{2}.$
C. $S=35.$
D. $S=\dfrac{75}{2}.$
& {{u}_{1}}=-3 \\
& {{u}_{n+1}}={{u}_{n}}+\dfrac{5}{2},\ n\ge 1 \\
\end{aligned} \right. $. Tính $ S={{u}_{20}}-{{u}_{6}}$.
A. $S=33.$
B. $S=\dfrac{69}{2}.$
C. $S=35.$
D. $S=\dfrac{75}{2}.$
Do ${{u}_{n+1}}={{u}_{n}}+\dfrac{5}{2},\ n\ge 1\Rightarrow $ Dãy số trên là 1 CSC có ${{u}_{1}}=-3,d=\dfrac{5}{2}$
$\Rightarrow \left\{ \begin{aligned}
& {{u}_{20}}={{u}_{1}}+19d=-3+19.\dfrac{5}{2}=\dfrac{89}{2} \\
& {{u}_{6}}={{u}_{1}}+5a=-3+5.\dfrac{5}{2}=\dfrac{19}{2} \\
\end{aligned} \right.\Rightarrow S={{u}_{20}}-{{u}_{6}}=\dfrac{89}{2}-\dfrac{19}{2}=35$.
$\Rightarrow \left\{ \begin{aligned}
& {{u}_{20}}={{u}_{1}}+19d=-3+19.\dfrac{5}{2}=\dfrac{89}{2} \\
& {{u}_{6}}={{u}_{1}}+5a=-3+5.\dfrac{5}{2}=\dfrac{19}{2} \\
\end{aligned} \right.\Rightarrow S={{u}_{20}}-{{u}_{6}}=\dfrac{89}{2}-\dfrac{19}{2}=35$.
Đáp án C.