题目内容
已知数列{an},构造一个新数列a1,(a2-a1),(a3-a2),…,(an-an-1),…,此数列是首项为1,公比为
(1)求数列{an}的通项;
(2)求数列{an}的前n项和Sn.
【答案】分析:(1)因为新数列a1,(a2-a1),(a3-a2),…,(an-an-1),…,此数列是首项为1,公比为
的等比数列,根据等比数列的通项公式可得数列{an}的通项;
(2)根据等比数列的求和公式得到即可.
解答:解:(1)由题意an=a1+(a2-a1)+(a3-a2)+…+(an-an-1)=
=
[1-(
)n].
(2)Sn=
[n-(
+
+
+…+
)]=
[n-
(1-
)]=
n-
+
.
点评:考查学生对等比数列性质的掌握能力,以及数列求和和数列递推式的方法.

(2)根据等比数列的求和公式得到即可.
解答:解:(1)由题意an=a1+(a2-a1)+(a3-a2)+…+(an-an-1)=

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(2)Sn=

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点评:考查学生对等比数列性质的掌握能力,以及数列求和和数列递推式的方法.

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