题目内容
(本小题满分13分)设等差数列
满足
.
(1)求
的通项公式;
(2)求数列
的前n项和Sn及使得Sn最大的序号n的值.
【答案】
解:(1) 由
得
解得 ![]()
∴
··························································································· 6分
(2) 由 (1) 知,
∴当n = 5时,Sn取得最大·········································································· 13分
【解析】略
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