广西2020版高考数学一轮复习 高考大题专项练三 高考中的数列 文 6页

高考大题专项练三　高考中的数列1.(2018全国Ⅲ,文17)等比数列{an}中,a1=1,a5=4a3.(1)求{an}的通项公式;(2)记Sn为{an}的前n项和,若Sm=63,求m.解(1)设{an}的公比为q,由题设得an=qn-1.由已知得q4=4q2,解得q=0(舍去),q=-2或q=2.故an=(-2)n-1或an=2n-1.(2)若an=(-2)n-1,则Sn=1-(-2)n3.由Sm=63得(-2)m=-188,此方程没有正整数解.若an=2n-1,则Sn=2n-1.由Sm=63得2m=64,解得m=6.综上,m=6.2.设数列{an}的前n项和为Sn,已知a1=3,Sn+1=3Sn+3.(1)求数列{an}的通项公式;(2)若bn=nan+1-an,求数列{bn}的前n项和Tn.解(1)(方法一)∵Sn+1=3Sn+3,∴Sn+1+32=3Sn+32.∴Sn+32=S1+323n-1=92×3n-1=3n+12.∴当n≥2时,an=Sn-Sn-1=3n+12-3n2=3n,a1也适合.∴an=3n.(方法二)由Sn+1=3Sn+3(n∈N*),可知当n≥2时,Sn=3Sn-1+3,两式相减,得an+1=3an(n≥2).又a1=3,代入Sn+1=3Sn+3,得a2=9,故an=3n.\n(2)∵bn=nan+1-an=n3n+1-3n=12·n3n,∴Tn=1213+232+333+…+n3n,①∴13Tn=12132+233+334+…+n-13n+n3n+1,②由①-②,得23Tn=1213+132+133+134+…+13n-n3n+1,解得Tn=38-2n+38·3n.3.已知数列{an}的前n项和为Sn,且Sn=2an-1;数列{bn}满足bn-1-bn=bnbn-1(n≥2,n∈N*),b1=1.(1)求数列{an},{bn}的通项公式;(2)求数列anbn的前n项和Tn.解(1)由Sn=2an-1,得S1=a1=2a1-1,故a1=1.又Sn=2an-1,Sn-1=2an-1-1(n≥2),两式相减,得Sn-Sn-1=2an-2an-1,即an=2an-2an-1.故an=2an-1,n≥2.所以数列{an}是首项为1,公比为2的等比数列.故an=1·2n-1=2n-1.由bn-1-bn=bnbn-1(n≥2,n∈N*),得1bn-1bn-1=1.又b1=1,∴数列1bn是首项为1,公差为1的等差数列.∴1bn=1+(n-1)·1=n.∴bn=1n.(2)由(1)得anbn=n·2n-1.∴Tn=1·20+2·21+…+n·2n-1,∴2Tn=1·21+2·22+…+n·2n.\n两式相减,得-Tn=1+21+…+2n-1-n·2n=1-2n1-2-n·2n=-1+2n-n·2n.∴Tn=(n-1)·2n+1.4.(2018天津,文18)设{an}是等差数列,其前n项和为Sn(n∈N*);{bn}是等比数列,公比大于0,其前n项和为Tn(n∈N*).已知b1=1,b3=b2+2,b4=a3+a5,b5=a4+2a6.(1)求Sn和Tn;(2)若Sn+(T1+T2+…+Tn)=an+4bn,求正整数n的值.解(1)设等比数列{bn}的公比为q.由b1=1,b3=b2+2,可得q2-q-2=0.因为q>0,可得q=2,故bn=2n-1.所以,Tn=1-2n1-2=2n-1.设等差数列{an}的公差为d.由b4=a3+a5,可得a1+3d=4.由b5=a4+2a6,可得3a1+13d=16,从而a1=1,d=1,故an=n.所以,Sn=n(n+1)2.(2)由(1),有T1+T2+…+Tn=(21+22+…+2n)-n=2×(1-2n)1-2-n=2n+1-n-2.由Sn+(T1+T2+…+Tn)=an+4bn可得,n(n+1)2+2n+1-n-2=n+2n+1,整理得n2-3n-4=0,解得n=-1(舍),或n=4.所以,n的值为4.5.已知f(x)=2sinπ2x,集合M={x||f(x)|=2,x>0},把M中的元素从小到大依次排成一列,得到数列{an},n∈N*.(1)求数列{an}的通项公式;(2)记bn=1an+12,设数列{bn}的前n项和为Tn,求证:Tn<14.(1)解f(x)=2sinπ2x,集合M={x||f(x)|=2,x>0},\n则π2x=kπ+π2,解得x=2k+1(k∈Z),把M中的元素从小到大依次排成一列,得到数列{an},所以an=2n-1.(2)证明bn=1an+12=1(2n+1)2<14n2+4n=141n-1n+1,故Tn=b1+b2+…+bn<141-12+12-13+…+1n-1n+1=141-1n+1<14.6.(2018浙江,20)已知等比数列{an}的公比q>1,且a3+a4+a5=28,a4+2是a3,a5的等差中项.数列{bn}满足b1=1,数列{(bn+1-bn)an}的前n项和为2n2+n.(1)求q的值;(2)求数列{bn}的通项公式.解(1)由a4+2是a3,a5的等差中项,得a3+a5=2a4+4,所以a3+a4+a5=3a4+4=28,解得a4=8.由a3+a5=20,得8q+1q=20,解得q=2或q=12,因为q>1,所以q=2.(2)设cn=(bn+1-bn)an,数列{cn}前n项和为Sn,由cn=S1,n=1,Sn-Sn-1,n≥2,解得cn=4n-1.由(1)可知an=2n-1,所以bn+1-bn=(4n-1)·12n-1.故bn-bn-1=(4n-5)·12n-2,n≥2,bn-b1=(bn-bn-1)+(bn-1-bn-2)+…+(b3-b2)+(b2-b1)=(4n-5)·12n-2+(4n-9)·12n-3+…+7·12+3.\n设Tn=3+7·12+11·122+…+(4n-5)·12n-2,n≥2,12Tn=3·12+7·122+…+(4n-9)·12n-2+(4n-5)·12n-1,所以12Tn=3+4·12+4·122+…+4·12n-2-(4n-5)·12n-1,因此Tn=14-(4n+3)·12n-2,n≥2,又b1=1,所以bn=15-(4n+3)·12n-2.7.已知正项数列{an}的首项a1=1,前n项和Sn满足an=Sn+Sn-1(n≥2).(1)求证:{Sn}为等差数列,并求数列{an}的通项公式;(2)记数列1anan+1的前n项和为Tn,若对任意的n∈N*,不等式4Tn