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2019-2020年高中数学第一章数列1.3.2.2数列求和及应用课后演练提升北师大版必修一、选择题(每小题5分,共20分)1数列9,99,999,9 999的前n项和等于()A10n1B.(10n1)nC.(10n1) D.(10n1)n解析:an10n1Sna1a2an(101)(1021)(10n1)(1010210n)nn.答案:B2数列1,的前n项和Sn等于()A. B.C. D.解析:an2,所以Sn22.答案:B3已知数列an的通项an2n1,由bn所确定的数列bn的前n项之和是()An(n2) B.n(n4)C.n(n5) D.n(n7)解析:a1a2an(2n4)n22n.bnn2,bn的前n项和Sn.答案:C4设数列1,(12),(124),(12222n1)的前m项和为2 036,则m的值为()A8 B9C10 D11解析:an2n1,Sn2n1n2,代入选项检验,即得m10.答案:C二、填空题(每小题5分,共10分)5在数列an中,a11,a22,且an2an1(1)n(nN),则S100_.解析:当n为奇数时,an2an0,故奇数项为常数列1当n为偶数时,an2an2,故偶数项为等差数列S1005012 600.答案:2 6006若数列an的通项公式an,则前n项和Sn_.解析:an,Sn().答案:三、解答题(每小题10分,共20分)7(1)已知等差数列ana1,an,Sn5,求n和d;a14,S8172,求a8和d.(2)在等比数列an中,S230,S3155,求Sn;a1a310,a4a6,求S5.解析:(1)由题意,得5,解得n15.因为a15(151)d,所以d.由已知,得172,解得a839.因为a84(81)d39,所以d5.(2)由题意知解得或从而Sn5n1或Sn.方法一:由题意知解得从而S5.方法二:由(a1a3)q3a4a6,得q3.从而q.又a1a3a1(1q2)10,所以a18,从而S5.8求数列的前n项和Sn.解析:由an得,Sn123(n1)n,则Sn123(n1)n得,Snnn1,Sn22.9(10分)设数列an满足a12,an1an322n1.(1)求数列an的通项公式;(2)令bnnan,求数列bn的前n项和Sn.解析:(1)由已知,当n1时,an1(an1an)(anan1)(a2a1)a13(22n122n32)222(n1)1.而a12,所以数列an的通项公式为an22n1.(2)由bnnann22n1知Sn12223325n22n1,从而22Sn123225327n22n1,得(122)Sn2232522n1n22n1,即Sn(3n1)22n12
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