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高中数学选择性必修 第二册(381题)


设 \({S_n}\) 是数列 \(\left\{ {{a_n}} \right\}\) 的前 \(n\) 项和,若 \({a_1} = \frac{1}{2}\) , \({a_{n + 1}} = 1 - \frac{1}{{{a_n}}}\) ,则 \({S_{2021}} = \)(     )


A.\(\frac{{2017}}{2}\)

B.\(1009\)

C.\(\frac{{2019}}{2}\)

D.\(1010\)


知识点:第四章 数列


参考答案:B


解析:

在数列 \(\left\{ {{a_n}} \right\}\)中,\({a_1} = \frac{1}{2}\)\({a_{n + 1}} = 1 - \frac{1}{{{a_n}}}\),则 \({a_2} = 1 - \frac{1}{{{a_1}}} = - 1\)\({a_3} = 1 - \frac{1}{{{a_2}}} = 2\)\({a_4} = 1 - \frac{1}{{{a_3}}} = \frac{1}{2}\),以此类推可知,对任意的 \(n \in {N^ * }\)\({a_{n + 3}} = {a_n}\),即数列 \(\left\{ {{a_n}} \right\}\) 是以\(3\)为周期的周期数列,

\(\because 2021 = 3 \times 673 + 2\), 因此,\({S_{2021}} = 673{S_3} + {a_1} + {a_2} = 674{S_3} - {a_3} = 674 \times \left( {\frac{1}{2} - 1 + 2} \right) - 2 = 1009\).

故选:B.

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