【答案】(1)设{an}的公差为d,则Sn=na1?n(n?1)d. 2?3a1?3d?0,解得a1?1,d??1.?由已知可得?5a1?10d??5,
故?an?的通项公式为an=2-n.
(2)由(I)知
11111??(?),
a2n?1a2n?1(3?2n)(1?2n)22n?32n?1从而数列?
??1111111n1(-+-+?+?)?. 的前n项和为?2-11132n?32n?11?2n?a2n?1a2n?1?
【答案】(1)设{an}的公差为d,则Sn=na1?n(n?1)d. 2?3a1?3d?0,解得a1?1,d??1.?由已知可得?5a1?10d??5,
故?an?的通项公式为an=2-n.
(2)由(I)知
11111??(?),
a2n?1a2n?1(3?2n)(1?2n)22n?32n?1从而数列?
??1111111n1(-+-+?+?)?. 的前n项和为?2-11132n?32n?11?2n?a2n?1a2n?1?
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