1、增分强化练(十五)考点一利用递推关系或Sn、an的关系求an1(2019晋城模拟)数列an满足a13,且对于任意的nN*都有an1ann2,则a39_.解析:因为an1ann2,所以a2a13,a3a24,a4a35,anan1n1(n2),上面n1个式子左右两边分别相加得ana1,即an,所以a39820.答案:8202(2019宝鸡模拟)若数列an满足a12a24a32n1an8n(nN*),则an_.解析:当n1时,a18.因为a12a24a32n1an8n,所以a12a24a32n2an18n8,(n2)两式相减得2n1an823,所以an24n(n2),适合n1.所以an24n.答案
2、:24n考点二数列求和1(2019安阳模拟)已知各项为正的等比数列an的前n项和为Sn,a13,且a2,15,S3依次成等差数列(1)求an;(2)若bn102n,求数列anbn的前n项和Tn.解析:(1)设各项为正的公比为q的等比数列an的前n项和为Sn,a13,且a2,15,S3依次成等差数列所以S3a230,即a1a1q230解得q3或3(负值舍去)故an33n13n.(2)由于bn102n,则anbn3n102n,所以Tn(31323n)(86102n)n29n.2(2019湛江模拟)Sn为数列an的前n项和,已知Snn2n.(1)求an的通项公式;(2)设bn,Tnb1b2bn,求T
3、n.解析:(1)当n2时,anSnSn1n2n(n1)2(n1)n11,当n1时,满足上式,ann11.(2)由ann11,可得bn,Tn.3(2019汕头模拟)记Sn为数列an的前n项和,若a119,Snnan1n(n1)(1)求数列an的通项公式;(2)设bn|an|,设数列bn的前n项和为Tn,求T20的值解析:(1)当n2时,因为Snnan1n(n1),所以Sn1(n1)ann(n1),得:annan1(n1)an2n,即an1an2(n2),又S1a22即a2a12,所以数列an是以19为首项,2为公差的等差数列,所以an19(n1)(2)212n.(2)由(1)知an212n,所以bn|an|212n|,因为当n10时an0,当n10时an0,所以bn,所以T20b1b2b20(19171)(1319)2(19171)2200.考点三数列的应用与综合问题(2019三明质检)已知数列an的前n项和为Sn,且a12,an1Sn2.(1)求数列an的前n项和Sn;(2)设bnlog2(S3n2),数列的前n项和为Tn,求证4Tn0,所以Tn.又Tn,当n1时,Tn取得最小值为,所以Tn,即4Tn.