2021年高考数学乙卷-理19 |
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2022-05-03 08:09:20 |
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19.(12分)记Sn为数列{an}的前n项和,bn为数列{Sn}的前n项积,已知2Sn+1bn=2. (1)证明:数列{bn}是等差数列; (2)求{an}的通项公式. 分析:(1)由题意当n=1时,b1=S1,代入已知等式可得b1的值,当n⩾2时,将bnbn−1=Sn,代入2Sn+1bn=2,可得bn−bn−1=12,进一步得到数列{bn}是等差数列; (2)由a1=S1=b1=32,可得bn=n+22,代入已知等式可得Sn=n+2n+1,当n⩾2时,an=Sn−Sn−1=−1n(n+1),进一步得到数列{an}的通项公式. 解:(1)证明:当n=1时,b1=S1, 由2b1+1b1=2,解得b1=32, 当n⩾2时,bnbn−1=Sn,代入2Sn+1bn=2, 消去Sn,可得2bn−1bn+1bn=2,所以bn−bn−1=12, 所以{bn}是以32为首项,12为公差的等差数列. (2)由题意,得a1=S1=b1=32, 由(1),可得bn=32+(n−1)×12=n+22, 由2Sn+1bn=2,可得Sn=n+2n+1, 当n⩾2时,an=Sn−Sn−1=n+2n+1−n+1n=−1n(n+1),显然a1不满足该式, 所以an={32,n=1−1n(n+1),n⩾2. 点评:本题考查了等差数列的概念,性质和通项公式,考查了方程思想,是基础题.
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