试题与答案

对于数列{an}，定义数列{an＋1－an}为数列{an}的“差数列”，若a1＝

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题型：填空题

题目：

对于数列{an}，定义数列{an＋1－an}为数列{an}的“差数列”，若a1＝1.{an}的“差数列”的通项公式为an＋1－an＝2n，则数列{an}的前n项和Sn＝________.

答案：

2n＋1－n－2

因为an＋1－an＝2n，应用累加法可得an＝2n－1，所以Sn＝a1＋a2＋a3＋…＋an＝2＋22＋23＋…＋2n－n＝－n＝2n＋1－n－2.

考点：等比数列的定义及性质
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