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1、(四)数列1(2017全国)已知等差数列an的前n项和为Sn,等比数列bn的前n项和为Tn,a11,b11,a2b22.(1)若a3b35,求bn的通项公式;(2)若T321,求S3.解设an的公差为d,bn的公比为q,则an1(n1)d,bnqn1.由a2b22,得dq3.(1)由a3b35,得2dq26.联立和,解得(舍去),因此bn的通项公式bn2n1.(2)由b11,T321,得q2q200.解得q5或q4.当q5时,由得d8,则S321.当q4时,由得d1,则S36.2(2017河北省衡水中学二模)已知数列an满足a11,2,nN*.(1)求数列an的通项公式;(2)设以2为公比的等
2、比数列bn满足4log2bnlog2bn1an12n11(nN*),求数列bnlog2bn的前n项和Sn.解(1)由题意知,数列是以2为首项,2为公差的等差数列,22(n1)2n,故an4n23.(2)设等比数列bn的首项为b1,则bnb12n1,依题意有4log2bnlog2bn14log2(b12n1)log2(b12n)4(log2b1n1)(log2b1n)4(log2b1)24log2b14(2log2b11)n4n24n212n8,即解得log2b12,b14,故bn42n12n1.bnlog2bn2n1(n1),Sn2n24.3(2017届辽宁省锦州市质检)已知等比数列an的前n
3、项和为Sn,a11,S69S3.(1)求an的通项公式;(2)设bn1log2an,求数列anbn的前n项和解(1)设等比数列an的公比为q,由a11,S69S3知,q1,故,即(1q3)(1q3)9(1q3),即1q39,即q38,解得q2,则ana1qn12n1.(2)bn1log2an1log22n11n1n,anbnn2n1,Tn122322n2n1,2Tn12222(n1)2n1n2n,由,得Tnn2n(12222n1)n2n(2n1)(n1)2n1.4(2017届湖南省长沙市雅礼中学模拟)已知在数列an中,a11,a39,且anan1n1(n2,nN*)(1)求的值及数列an的通项
4、公式;(2)设bn(1)n(ann),且数列bn的前2n项和为S2n,求S2n.解(1)a11,a39,且anan1n1(n2,nN*),a22,a3519,解得2,anan12n1(n2,nN*)an(2n1)(2n3)31n2.(2)bn(1)n(ann)(1)n(n2n),b2n1b2n(2n1)2(2n1)(2n)22n4n,S2n42n22n.5(2017湖南省衡阳市联考)已知在数列an中,a12,anan12n0(n2,nN*)(1)写出a2,a3的值(只写出结果),并求出数列an的通项公式;(2)设bn,若对任意的正整数n,不等式t22tbn恒成立,求实数t的取值范围解(1)a26,a312,当n2时,ana1(a2a1)(a3a2)(anan1)2(123n)n(n1),当n1时,a12也满足上式,ann(n1)(2)bn.bn1bn0,bn1bnt22tt22t0t2,t(,0)(2,)