2022新高考總復(fù)習(xí)《數(shù)學(xué)》(人教)高考大題專項(xiàng)練高考大題專項(xiàng)(三)　數(shù)列

1、 高考大題專項(xiàng)(三)數(shù)列1.(2020陜西咸陽(yáng)高三第一次模擬)已知數(shù)列an的前n項(xiàng)和為Sn,且滿足Sn=2an-2n-1(nN*).(1)求證:數(shù)列an+2是等比數(shù)列;(2)求數(shù)列n(an+2)的前n項(xiàng)和.2.(2020山東高考預(yù)測(cè)卷)在b4=a3+a5;b4+b6=3a3+3a5;a2+a3=b4這三個(gè)條件中任選一個(gè),補(bǔ)充在下面的問(wèn)題中,若問(wèn)題中的k存在,求出k的值;若k不存在,說(shuō)明理由.已知an是等差數(shù)列,其前n項(xiàng)和為Sn,bn是公比大于0的等比數(shù)列,b1=1,b3=b2+2,b5=a4+2a6,且,設(shè)cn=b2Sn,是否存在k,使得對(duì)任意的nN*,都有ckcn?3.若數(shù)列an的前n項(xiàng)和為

2、Sn,且a1=1,a2=2,(Sn+1)(Sn+2+1)=(Sn+1+1)2.(1)求Sn;(2)記數(shù)列1an的前n項(xiàng)和為Tn,證明:1Tn2.4.(2020江西師大附中、鷹潭一中高三?？?在數(shù)列an中,a1=1,a1+a22+a33+ann=2n-1(nN*).(1)求數(shù)列an的前n項(xiàng)和Sn;(2)若存在nN*,使得ann(n+1)成立,求實(shí)數(shù)的最小值.5.(2020安徽合肥高三第二次質(zhì)檢)已知等差數(shù)列an的前n項(xiàng)和為Sn,a2=1,S7=14,數(shù)列bn滿足b1b2b3bn=2n2+n2.(1)求數(shù)列an和bn的通項(xiàng)公式;(2)若數(shù)列cn滿足cn=bncos(an),求數(shù)列cn的前2n項(xiàng)和T

3、2n.6.(2020天津,19)已知an為等差數(shù)列,bn為等比數(shù)列,a1=b1=1,a5=5(a4-a3),b5=4(b4-b3).(1)求an和bn的通項(xiàng)公式;(2)記an的前n項(xiàng)和為Sn,求證:SnSn+20),因?yàn)閎n是公比大于0的等比數(shù)列,且b1=1,b3=b2+2,所以q2=q+2,解得q=2(q=-1不合題意,舍去).所以bn=2n-1.若存在k,使得對(duì)任意的nN*,都有ckcn,則cn存在最小值.方案一:若選,則由b5=a4+2a6,b4=a3+a5,可得3a1+13d=16,2a1+6d=8,解得a1=1,d=1.所以Sn=12n2+12n,cn=b2Sn=212n2+12n=

4、4n2+n.因?yàn)閚N*,所以n2+n2,cn是遞減數(shù)列,所以cn不存在最小值,即不存在滿足題意的k.方案二:若選,由b5=a4+2a6,b4+b6=3a3+3a5,可得3a1+13d=16,6a1+18d=40,解得a1=293,d=-1.所以Sn=-12n2+616n,cn=b2Sn=12-3n2+61n.因?yàn)楫?dāng)n20時(shí),cn0,當(dāng)n21時(shí),cn0,所以易知cn的最小值為c21=-27.即存在k=21,使得對(duì)任意的nN*,都有ckcn.方案三:若選,則由b5=a4+2a6,a2+a3=b4,可得3a1+13d=16,2a1+3d=8,解得a1=5617,d=817.所以Sn=4n2+52n1

5、7,cn=b2Sn=172n2+26n.因?yàn)閚N*,所以2n2+26n28,cn是遞減數(shù)列,所以cn不存在最小值,即不存在滿足題意的k.3.(1)解由題意得Sn+2+1Sn+1+1=Sn+1+1Sn+1=S2+1S1+1,所以數(shù)列Sn+1是等比數(shù)列.又因?yàn)镾1+1=a1+1=2,S2+1=a1+a2+1=4,所以S2+1S1+1=2,所以數(shù)列Sn+1是首項(xiàng)為2,公比為2的等比數(shù)列.所以Sn+1=22n-1=2n,所以Sn=2n-1.(2)證明由(1)知,當(dāng)n2時(shí),Sn=2n-1,Sn-1=2n-1-1,兩式相減得an=2n-1.當(dāng)n=1時(shí),a1=1也滿足an=2n-1,所以數(shù)列an的通項(xiàng)公式為

6、an=2n-1(nN*).所以1an=12n-1(nN*).所以Tn=1a1+1a2+1an=1+12+12n-1=1-(12)n1-12=2-12n-1.因?yàn)閚N*,所以012n-11,所以-1-12n-10.所以12-12n-12,即1Tn1,f(n)單調(diào)遞增,從而f(n)min=f(1)=12,12,因此實(shí)數(shù)的最小值為12.5.解(1)設(shè)an的公差為d,由a2=1,S7=14得a1+d=1,7a1+21d=14.解得a1=12,d=12,an=n2.b1b2b3bn=2n2+n2=2n(n+1)2,b1b2b3bn-1=2n(n-1)2(n2),兩式相除得bn=2n(n2).當(dāng)n=1時(shí),

7、b1=2符合上式.bn=2n.(2)cn=bncos(an)=2ncosn2,T2n=2cos2+22cos +23cos32+24cos(2)+22n-1cos(2n-1)2+22ncos(n)=22cos +24cos(2)+22ncos(n)=-22+24-+(-1)n22n=-41-(-4)n1+4=-4+(-4)n+15.6.(1)解設(shè)等差數(shù)列an的公差為d,等比數(shù)列bn的公比為q.由a1=1,a5=5(a4-a3),可得d=1,從而an的通項(xiàng)公式為an=n.由b1=1,b5=4(b4-b3),又q0,可得q2-4q+4=0,解得q=2,從而bn的通項(xiàng)公式為bn=2n-1.(2)證明

8、由(1)可得Sn=n(n+1)2,故SnSn+2=14n(n+1)(n+2)(n+3),Sn+12=14(n+1)2(n+2)2,從而SnSn+2-Sn+12=-12(n+1)(n+2)0,所以SnSn+2Sn+12.(3)解當(dāng)n為奇數(shù)時(shí),cn=(3an-2)bnanan+2=(3n-2)2n-1n(n+2)=2n+1n+2-2n-1n;當(dāng)n為偶數(shù)時(shí),cn=an-1bn+1=n-12n.對(duì)任意的正整數(shù)n,有k=1nc2k-1=k=1n22k2k+1-22k-22k-1=22n2n+1-1,和k=1nc2k=k=1n2k-14k=14+342+543+2n-14n.由得14k=1nc2k=142+343+2n-34n+2n-14n+1.由得34k=1nc2k=14+242+24n-2n-14n