高考大題專項(xiàng)練三　高考中的數(shù)列 1.(2018全國(guó)Ⅲ,文17)等比數(shù)列{an}中,a1=1,a5=4a3. (1)求{an}的通項(xiàng)公式; (2)記Sn為{an}的前n項(xiàng)和,若Sm=63,求m. 解(1)設(shè){an}的公比為q,由題設(shè)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,則Sn=1-(-2)n3. 由Sm=63得(-2)m=-188,此方程沒有正整數(shù)解. 若an=2n-1,則Sn=2n-1.由Sm=63得2m=64,解得m=6. 綜上,m=6. 2.設(shè)數(shù)列{an}的前n項(xiàng)和為Sn,已知a1=3,Sn+1=3Sn+3. (1)求數(shù)列{an}的通項(xiàng)公式; (2)若bn=nan+1-an,求數(shù)列{bn}的前n項(xiàng)和Tn. 解(1)(方法一)∵Sn+1=3Sn+3, ∴Sn+1+32=3Sn+32. ∴Sn+32=S1+323n-1=923n-1=3n+12. ∴當(dāng)n≥2時(shí),an=Sn-Sn-1=3n+12-3n2=3n,a1也適合. ∴an=3n. (方法二)由Sn+1=3Sn+3(n∈N*), 可知當(dāng)n≥2時(shí),Sn=3Sn-1+3, 兩式相減,得an+1=3an(n≥2). 又a1=3,代入Sn+1=3Sn+3,得a2=9,故an=3n. (2)∵bn=nan+1-an=n3n+1-3n=12n3n, ∴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+383n. 3.已知數(shù)列{an}的前n項(xiàng)和為Sn,且Sn=2an-1;數(shù)列{bn}滿足bn-1-bn=bnbn-1(n≥2,n∈N*),b1=1. (1)求數(shù)列{an},{bn}的通項(xiàng)公式; (2)求數(shù)列anbn的前n項(xiàng)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), 兩式相減,得Sn-Sn-1=2an-2an-1, 即an=2an-2an-1. 故an=2an-1,n≥2. 所以數(shù)列{an}是首項(xiàng)為1,公比為2的等比數(shù)列. 故an=12n-1=2n-1. 由bn-1-bn=bnbn-1(n≥2,n∈N*), 得1bn-1bn-1=1. 又b1=1,∴數(shù)列1bn是首項(xiàng)為1,公差為1的等差數(shù)列. ∴1bn=1+(n-1)1=n.∴bn=1n. (2)由(1)得anbn=n2n-1. ∴Tn=120+221+…+n2n-1, ∴2Tn=121+222+…+n2n. 兩式相減,得-Tn=1+21+…+2n-1-n2n =1-2n1-2-n2n=-1+2n-n2n. ∴Tn=(n-1)2n+1. 4.(2018天津,文18)設(shè){an}是等差數(shù)列,其前n項(xiàng)和為Sn(n∈N*);{bn}是等比數(shù)列,公比大于0,其前n項(xiàng)和為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,求正整數(shù)n的值. 解(1)設(shè)等比數(shù)列{bn}的公比為q.由b1=1,b3=b2+2, 可得q2-q-2=0.因?yàn)閝>0,可得q=2,故bn=2n-1. 所以,Tn=1-2n1-2=2n-1. 設(shè)等差數(shù)列{an}的公差為d.由b4=a3+a5,可得a1+3d=4. 由b5=a4+2a6,可得3a1+13d=16, 從而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的值為4. 5.已知f(x)=2sinπ2x,集合M={x||f(x)|=2,x>0},把M中的元素從小到大依次排成一列,得到數(shù)列{an},n∈N*. (1)求數(shù)列{an}的通項(xiàng)公式; (2)記bn=1an+12,設(shè)數(shù)列{bn}的前n項(xiàng)和為Tn,求證:Tn<14. (1)解f(x)=2sinπ2x,集合M={x||f(x)|=2,x>0}, 則π2x=kπ+π2,解得x=2k+1(k∈Z), 把M中的元素從小到大依次排成一列,得到數(shù)列{an}, 所以an=2n-1. (2)證明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)已知等比數(shù)列{an}的公比q>1,且a3+a4+a5=28,a4+2是a3,a5的等差中項(xiàng).數(shù)列{bn}滿足b1=1,數(shù)列{(bn+1-bn)an}的前n項(xiàng)和為2n2+n. (1)求q的值; (2)求數(shù)列{bn}的通項(xiàng)公式. 解(1)由a4+2是a3,a5的等差中項(xiàng),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?yàn)閝>1,所以q=2. (2)設(shè)cn=(bn+1-bn)an,數(shù)列{cn}前n項(xiàng)和為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+…+712+3. 設(shè)Tn=3+712+11122+…+(4n-5)12n-2,n≥2, 12Tn=312+7122+…+(4n-9)12n-2+(4n-5)12n-1, 所以12Tn=3+412+4122+…+412n-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.已知正項(xiàng)數(shù)列{an}的首項(xiàng)a1=1,前n項(xiàng)和Sn滿足an=Sn+Sn-1(n≥2). (1)求證:{Sn}為等差數(shù)列,并求數(shù)列{an}的通項(xiàng)公式; (2)記數(shù)列1anan+1的前n項(xiàng)和為Tn,若對(duì)任意的n∈N*,不等式4Tn<a2-a恒成立,求實(shí)數(shù)a的取值范圍. 解(1)因?yàn)閍n=Sn+Sn-1, 所以Sn-Sn-1=Sn+Sn-1, 即Sn-Sn-1=1, 所以數(shù)列{Sn}是首項(xiàng)為S1=a1=1,公差為1的等差數(shù)列,得Sn=n, 所以an=Sn+Sn-1=n+(n-1)=2n-1(n≥2), 當(dāng)n=1時(shí),a1=1也適合,所以an=2n-1. (2)因?yàn)?anan+1=1(2n-1)(2n+1) =1212n-1-12n+1, 所以Tn=121-13+13-15+…+12n-1-12n+1 =121-12n+1. 所以Tn<12.要使不等式4Tn<a2-a恒成立,只需2≤a2-a恒成立,解得a≤-1或a≥2, 故實(shí)數(shù)a的取值范圍是(-∞,-1]∪[2,+∞). 8.已知數(shù)列{an}是公比為12的等比數(shù)列,其前n項(xiàng)和為Sn,且1-a2是a1與1+a3的等比中項(xiàng),數(shù)列{bn}是等差數(shù)列,其前n項(xiàng)和Tn滿足Tn=nλbn+1(λ為常數(shù),且λ≠1),其中b1=8. (1)求數(shù)列{an}的通項(xiàng)公式及λ的值; (2)比較1T1+1T2+1T3+…+1Tn與12Sn的大小. 解(1)由題意,得(1-a2)2=a1(a3+1), 即1-12a12=a114a1+1, 解得a1=12.故an=12n. 設(shè)等差數(shù)列{bn}的公差為d, 又T1=λb2,T2=2λb3,即8=λ(8+d),16+d=2λ(8+2d), 解得λ=12,d=8或λ=1,d=0(舍去),故λ=12. (2)由(1)知Sn=1-12n, 則12Sn=12-12n+1≥14. ① 由(1)知Tn=12nbn+1,當(dāng)n=1時(shí),T1=b1=12b2, 即b2=2b1=16,故公差d=b2-b1=8, 則bn=8n,又Tn=nλbn+1, 故Tn=4n2+4n,即1Tn=14n(n+1)=141n-1n+1. 因此,1T1+1T2+…+1Tn =141-12+12-13+…+1n-1n+1 =141-1n+1<14. ② 由①②可知1T1+1T2+…+1Tn<12Sn.