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文（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)若数列{bn}的前n项和为Tn,求数列{Tn}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;②=3;③-3an+1=+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山东淄博一模)已知数列{an}中,a1=1,an+1=2an+3×2n-1(n∈N*).(1)判断数列是否为等差数列,并说明理由;(2)求数列{an}的前n项和Sn. 5.(2023山东潍坊一模)已知数列{an}为等比数列,其前n项和为Sn,且满足Sn=2n+m(m∈R).(1)求m的值及数列{an}的通项公式;(2)设bn=|log2an-5|,求数列{bn}的前n项和Tn.6.(2023新高考Ⅱ,18)已知{an}为等差数列,bn=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an+2=an+1+2an,得bn+1=an+2+an+1=2(an+1+an)=2bn.又b1=a1+a2=2≠0,所以{bn}是以2为首项,2为公比的等比数列.(2)解由(1)知,Tn==2n+1-2.设数列{Tn}的前n项和为Sn,由Tn=2n+1-2,知Sn=(22+23+…+2n+1)-2n=-2n=2n+2-2n-4.2.(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,∴,∴Tn=+…+=1-.3.解(1)若选择条件①.因为a1,a3,a11成等比数列,所以=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=3,所以=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-3an+1=+3an,所以=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+×6+5n+×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)数列是等差数列,理由如下:因为,所以数列是以为首项,以为公差的等差数列.(2)由(1)知数列的通项公式为+(n-1)×(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×-(3n-1)×2n-1=-2+(4-3n)·2n-1,则Sn=2+(3n-4)·2n-1.5.解(1)因为Sn=2n+m,当n≥2时,Sn-1=2n-1+m,两式相减得an=2n-1.又由数列{an}为等比数列,则n=1时满足an=2n-1,所以a1=S1=21+m=21-1=1,则m=-1.{an}的通项公式为an=2n-1.(2)由(1)知bn=|log2an-5|=|n-6|,当1≤n≤6时,bn=6-n,Tn=,当n>6时,Tn=T6+×(n-6)=15+.综上,Tn=6.(1)解设等差数列{an}的公差为d.由bn=得b1=a1-6,b2=2a2=2(a1+d),b3=a3-6=a1+2d-6.则由S4=32,T3=16,得 解得所以an=a1+(n-1)d=2n+3.(2)证明由(1)可得Sn==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)]=.当n>5时,Tn-Sn=-(n2+4n)=>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)]=.当n>5时,Tn-Sn=-(n2+4n)=>0,所以Tn>Sn.综上可知,当n>5时,Tn>Sn.