1、单元质检六数列(B)(时间:45分钟满分:100分)一、选择题(本大题共6小题,每小题7分,共42分)1.在单调递减的等比数列an中,若a3=1,a2+a4=52,则a1=()A.2B.4C.2D.22答案:B解析:设an的公比为q.由已知,得a1q2=1,a1q+a1q3=52,q+q3q2=52,q2-52q+1=0,q=12(q=2舍去),a1=4.2.设an=-n2+9n+10,则数列an前n项和最大时n的值为()A.9B.10C.9或10D.12答案:C解析:令an0,得n2-9n-100,1n10.令an+10,即n2-7n-180,n9.9n10.前9项和等于前10项和,它们都最
2、大.3.公差不为零的等差数列an的前n项和为Sn.若a4是a3与a7的等比中项,S8=16,则S10等于()A.18B.24C.30D.60答案:C解析:设等差数列an的公差为d0.由题意,得(a1+3d)2=(a1+2d)(a1+6d),化为2a1+3d=0,S8=16,8a1+872d=16,联立解得a1=-32,d=1.则S10=10-32+10921=30.4.在数列an中,a1=1,an+1=2an,Sn为an的前n项和.若Sn+为等比数列,则=()A.-1B.1C.-2D.2答案:B解析:由题意,得an是等比数列,公比为2,Sn=2n-1,Sn+=2n-1+.Sn+为等比数列,-1
3、+=0,=1,故选B.5.设等差数列an的前n项和为Sn,若S6S7S5,则满足SnSn+1S7S5,6a1+652d7a1+762d5a1+542d,a70,S13=13(a1+a13)2=13a70,满足SnSn+10,两边取以2为底的对数可得log2(an+1)=log2(an-1+1)2=2log2(an-1+1),则数列log2(an+1)是以1为首项,2为公比的等比数列,log2(an+1)=2n-1,an=22n-1-1,又an=an-12+2an-1(n2),可得an+1=an2+2an(nN*),两边取倒数可得1an+1=1an2+2an=1an(an+2)=121an-1a
4、n+2,即2an+1=1an-1an+2,因此bn=1an+1+1an+2=1an-1an+1,所以Sn=b1+bn=1a1-1an+1=1-122n-1,故答案为1-122n-1.三、解答题(本大题共3小题,共44分)9.(14分)(2020山东,18)已知公比大于1的等比数列an满足a2+a4=20,a3=8.(1)求an的通项公式;(2)记bm为an在区间(0,m(mN*)中的项的个数,求数列bm的前100项和S100.解:(1)设an的公比为q.由题设得a1q+a1q3=20,a1q2=8.解得q=12(舍去),q=2.因为a1q2=8,所以a1=2.所以an的通项公式为an=2n.(
5、2)由题设及(1)知b1=0,且当2nm0,可得q=2,故an=2n-1.设等差数列bn的公差为d.由a4=b3+b5,可得b1+3d=4.由a5=b4+2b6,可得3b1+13d=16,从而b1=1,d=1,故bn=n.所以,数列an的通项公式为an=2n-1,数列bn的通项公式为bn=n.(2)解由(1),有Sn=1-2n1-2=2n-1,故Tn=k=1n(2k-1)=k=1n2k-n=2(1-2n)1-2-n=2n+1-n-2.证明因为(Tk+bk+2)bk(k+1)(k+2)=(2k+1-k-2+k+2)k(k+1)(k+2)=k2k+1(k+1)(k+2)=2k+2k+2-2k+1k+1,所以,k=1n(Tk+bk+2)bk(k+1)(k+2)=233-222+244-233+2n+2n+2-2n+1n+1=2n+2n+2-2.
