高考理科数学二轮总复习课后习题 考点突破练与专题检测5 数列求和方法及综合应用.docVIP

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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.(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{2nana

3.(云南红河二模)已知等差数列{an}的公差d0,a1=2,其前n项和为Sn,且.?

在①a1,a3,a11成等比数列;②S55-S33=3;③an+1

(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.(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{an2

(2)求数列{an}的前n项和Sn.

5.(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+3a

(1){an}的通项公式;

(2){bn}前n项和的最值.

6.(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S

(1)求{an}的通项公式;

(2)证明:当n5时,TnSn.

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an0,∴bn0.∵bn+1bn=an+2+an+1a

(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

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(

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d0,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(a

(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]

(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{an2

因为an+12n+1-an2n=

(2)由(1)知数列{an2n}的通项公式为an2n=12+(n-1)×

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-1,②

①-②得-Sn=1+3×(20+21+…+2n-2)-(3n-1)×2n-1=1+3×1-2n-11-2

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a

∴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×(1

∴Sn=(12+13)-(13+14)+…+(-1)n(1n+1

当n为奇数时,Sn=12+1n+2,{Sn}是递减数列,∴1

当n为偶数时,Sn=12-1n+2,{Sn}是递增数列,∴S2≤Sn12,∴{bn}前n项和的最大值为S1=

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

解得a1=5,d=2.

(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