题目

已知等差数列{an}的公差为2，前n项和为Sn ， 且S1 ， S2 ， S4成等比数列． （Ⅰ）求数列{an}的通项公式；（Ⅱ）令bn=（﹣1）n﹣1 ，求数列{bn}的前n项和Tn ． 答案：解：（Ⅰ）∵等差数列{an}的公差为2，前n项和为Sn， ∴Sn= na1+n(n−1)2d =n2﹣n+na1，∵S1，S2，S4成等比数列，∴ S22=S1⋅S4 ，∴ (22−2+2a1)2=a1⋅(42−4+4a1) ，化为 (1+a1)2=a1(3+a1) ，解得a1=1．∴an=a1+（n﹣1）d=1+2（n﹣1）=2n﹣1．（Ⅱ）由（Ⅰ）可得bn=（﹣1）n﹣1 4nanan+1 = (−1)n−1⋅4n(2n−1)(2n+1) = (−1)n−1(12n−1+12n+1) ．∴Tn= (1+13) ﹣ (13+15) + (15+17) ++ (−1)n−1(12n−1+12n+1) ．当n为偶数时，Tn= (1+13) ﹣ (13+15) + (15+17) ++ (12n−3+12n−1) ﹣ (12n−1+12n+1) =1﹣ 12n+1 = 2n2n+1 ．当n为奇数时，Tn= (1+13) ﹣ (13+15) + (15+17) +﹣ (12n−3+12n−1) + (12n−1+12n+1) =1+ 12n+1 = 2n+22n+1 ．∴Tn= {2n2n+1，n为偶数2n+22n+1，n为奇数 ．

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