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Câu hỏi: Cho cấp số cộng $\left( {{u}_{n}} \right)$ có ${{u}_{1}}=-\dfrac{1}{4}, d=\dfrac{1}{2}$. Dạng khai triển của cấp số cộng đó là
A. $-\dfrac{1}{4};\dfrac{1}{2};1;\dfrac{3}{2};...$
B. $-\dfrac{1}{4};\dfrac{1}{2};\dfrac{3}{2};\dfrac{5}{2};...$
C. $-\dfrac{1}{4};\dfrac{1}{4};\dfrac{1}{2};1;...$
D. $-\dfrac{1}{4};\dfrac{1}{4};\dfrac{3}{4};\dfrac{5}{4};...$
A. $-\dfrac{1}{4};\dfrac{1}{2};1;\dfrac{3}{2};...$
B. $-\dfrac{1}{4};\dfrac{1}{2};\dfrac{3}{2};\dfrac{5}{2};...$
C. $-\dfrac{1}{4};\dfrac{1}{4};\dfrac{1}{2};1;...$
D. $-\dfrac{1}{4};\dfrac{1}{4};\dfrac{3}{4};\dfrac{5}{4};...$
Ta có ${{u}_{2}}={{u}_{1}}+d=-\dfrac{1}{4}+\dfrac{1}{2}=\dfrac{1}{4}$.
${{u}_{3}}={{u}_{2}}+d=\dfrac{1}{4}+\dfrac{1}{2}=\dfrac{3}{4}, {{u}_{4}}={{u}_{3}}+d=\dfrac{3}{4}+\dfrac{1}{2}=\dfrac{5}{4}$.
${{u}_{3}}={{u}_{2}}+d=\dfrac{1}{4}+\dfrac{1}{2}=\dfrac{3}{4}, {{u}_{4}}={{u}_{3}}+d=\dfrac{3}{4}+\dfrac{1}{2}=\dfrac{5}{4}$.
Đáp án D.