适用于老高考旧教材2024版高考数学二轮复习考点突破练5数列求和方法及综合应用理（附解析）

考点突破练5　数列求和方法及综合应用1.已知数列{an}中,a1=a2=1,且an+2=an+1+2an.记bn=an+1+an.(1)求证:数列{bn}是等比数列;(2)求数列{bn+2n}的前n项和.2.(2023山西晋中二模)已知数列{an}满足a1=1,an+1=2an+1.(1)求证:数列{an+1}为等比数列;(2)求数列{2nanan+1}的前n项和Tn.3.(2023云南红河二模)已知等差数列{an}的公差d>0,a1=2,其前n项和为Sn,且　　　　　. 在①a1,a3,a11成等比数列;②S55-S33=3;③an+12-3an+1=an2+3an这三个条件中任选一个,补充在横线上,并回答下列问题.(1)求数列{an}的通项公式;(2)若数列{bn}满足bn=1+(-1)nan,求数列{bn}的前2n项和T2n. 4.(2023山东淄博高三一模)已知数列{an}中,a1=1,an+1=2an+3×2n-1(n∈N*).(1)判断数列{an2n}是否为等差数列,并说明理由;(2)求数列{an}的前n项和Sn.5.(2023江西九江二模)已知公差不为零的等差数列{an}中,a1+a5=8,且a2,a5,a11成等比数列,记bn=(-1)n+1·2n+3anan+1.求:(1){an}的通项公式;(2){bn}前n项和的最值.6.(2023新高考Ⅱ,18)已知{an}为等差数列,bn=an-6,n为奇数,2an,n为偶数.记Sn,Tn分别为数列{an},{bn}的前n项和,S4=32,T3=16.(1)求{an}的通项公式;(2)证明:当n>5时,Tn>Sn. 考点突破练5　数列求和方法及综合应用1.(1)证明由题可知an>0,∴bn>0.∵bn+1bn=an+2+an+1an+1+an=an+1+2an+an+1an+1+an=2(an+1+an)an+1+an=2,且b1=a1+a2=2,∴{bn}是以2为首项,2为公比的等比数列.(2)解由(1)知,bn=2n,则bn+2n=2n+2n.设{bn+2n}的前n项和为Sn,则Sn=2×(1-2n)1-2+n(2+2n)2=2n+1+n2+n-2.2.(1)证明∵an+1=2an+1,∴an+1+1=2(an+1).又a1=1,∴a1+1=2≠0,∴{an+1}是以2为首项,2为公比的等比数列.(2)解由(1)知an+1=2n,∴an=2n-1,∴2nanan+1=2n(2n-1)(2n+1-1)=12n-1-12n+1-1,∴Tn=12-1-122-1+122-1-123-1+…+12n-1-12n+1-1=1-12n+1-1.3.解(1)若选择条件①.因为a1,a3,a11成等比数列,所以a32=a1a11,即(a1+2d)2=a1×(a1+10d),整理得2d2=3a1d.又d>0,解得d=3,所以数列{an}的通项公式为an=a1+(n-1)d=3n-1.若选择条件②.因为S55-S33=3,所以5a1+10d5-3a1+3d3=3,解得d=3,所以数列{an}的通项公式为an=a1+(n-1)d=3n-1.若选择条件③.因为an+12-3an+1=an2+3an,所以an+12-an2=3an+1+3an,即(an+1-an)(an+1+an)=3(an+1+an).因为a1=2,d>0,所以an+1+an≠0,所以an+1-an=3=d,则数列{an}的通项公式为an=a1+(n-1)d=3n-1.(2)由(1)知an=3n-1.(方法一)T2n=b1+b2+…+b2n-1+b2n=(1-a1)+(a2+1)+…+(1-a2n-1)+(a2n+1)=-(a1+a3+…+a2n-1)+n+(a2+a4+…+a2n)+n=-[2n+n(n-1)2×6]+[5n+n(n-1)2×6]+2n=5n.(方法二)T2n=b1+b2+…+b2n-1+b2n=(1-a1)+(a2+1)+…+(1-a2n-1)+(a2n+1)=(a2-a1)+(a4-a3)+…+(a2n-a2n-1)+n+n=3n+2n=5n.4.解(1)数列{an2n}是等差数列,理由如下:因为an+12n+1-an2n=2an+3×2n-12n+1-an2n=34,所以数列{an2n}是以a12=12为首项,以34为公差的等差数列.(2)由(1)知数列{an2n}的通项公式为an2n=12+(n-1)×34=14(3n-1),则an=(3n-1)·2n-2(n∈N*),所以Sn=2×2-1+5×20+8×21+…+(3n-4)×2n-3+(3n-1)×2n-2,①所以2Sn=2×20+5×21+8×22+…+(3n-4)×2n-2+(3n-1)×2n-1,②①-②得-Sn=1+3×(20+21+…+2n-2)-(3n-1)×2n-1=1+3×1-2n-11-2-(3n-1)×2n-1=-2+(4-3n)·2n-1,则Sn=2+(3n-4)·2n-1.5.解(1)∵a1+a5=8,∴a3=a1+2d=4.∵a2,a5,a11成等比数列,∴a52=a2a11,即(a1+4d)2=(a1+d)(a1+10d),化简得a1d-2d2=0.由a1+2d=4,a1d-2d2=0,解得a1=2,d=1或a1=4,d=0(舍去),∴an=2+(n-1)×1=n+1.(2)由(1)可知bn=(-1)n+1·2n+3(n+1)(n+2)=(-1)n+1×(1n+1+1n+2).设{bn}的前n项和为Sn,∴Sn=(12+13)-(13+14)+…+(-1)n(1n+1n+1)+(-1)n+1(1n+1+1n+2)=12+(-1)n+11n+2.当n为奇数时,Sn=12+1n+2,{Sn}是递减数列,∴12<Sn≤S1;当n为偶数时,Sn=12-1n+2,{Sn}是递增数列,∴S2≤Sn<12,∴{bn}前n项和的最大值为S1=12+11+2=56,最小值为S2=12-12+2=14.6.(1)解设等差数列{an}的公差为d.由bn=an-6,n为奇数,2an,n为偶数,得b1=a1-6,b2=2a2=2(a1+d),b3=a3-6=a1+2d-6.则由S4=32,T3=16,得4a1+4×32×d=32,(a1-6)+2(a1+d)+(a1+2d-6)=16,解得a1=5,d=2.所以an=a1+(n-1)d=2n+3.(2)证明由(1)可得Sn=n[5+(2n+3)]2=n2+4n.当n为奇数时,Tn=a1-6+2a2+a3-6+2a4+a5-6+2a6+…+an-2-6+2an-1+an-6=(-1+14)+(3+22)+(7+30)+…+[(2n-7)+(4n+2)]+2n-3=[-1+3+…+(2n-7)+(2n- 3)]+[14+22+…+(4n+2)]=n+12(-1+2n-3)2+n-12(14+4n+2)2=3n2+5n-102.当n>5时,Tn-Sn=3n2+5n-102-(n2+4n)=n2-3n-102=(n-5)(n+2)2>0,所以Tn>Sn.当n为偶数时,Tn=a1-6+2a2+a3-6+2a4+a5-6+2a6+…+an-1-6+2an=(-1+14)+(3+22)+(7+30)+…+[(2n-5)+(4n+6)]=[-1+3+…+(2n-5)]+[14+22+…+(4n+6)]=n2(-1+2n-5)2+n2(14+4n+6)2=3n2+7n2.当n>5时,Tn-Sn=3n2+7n2-(n2+4n)=n2-n2=n(n-1)2>0,所以Tn>Sn.综上可知,当n>5时,Tn>Sn.