链接高考7数列的综合

1.(2022新高考Ⅱ,17,10分)已知{an}是等差数列,{bn}是公比为2的等比数列,且a2-b2=a3-b3=b4-a4.

(1)证明:a1=b1;

(2)求集合{k|bk=am+a1,1≤m≤500}中元素的个数.

解析(1)证明:设等差数列{an}的公差为d.

由a2-b2=a3-b3得a1+d-2b1=a1+2d-4b1,故d=2b1,①

由a3-b3=b4-a4得a1+2d-4b1=8b1-a1-3d,

故2a1+5d=12b1,②

由①②得2a1+10b1=12b1,即a1=b1.

(2)由(1)知d=2b1=2a1,由bk=am+a1,1≤m≤500得b1×2k-1=2a1+(m-1)d,即a1×2k-1=2a1+2(m-1)a1,其中a1≠0,∴2k-1=2m,即2k-2=m,∴1≤2k-2≤500,∴0≤k-2≤8,

∴2≤k≤10.

故集合{k|bk=am+a1,1≤m≤500}中元素的个数为9.

2.(2023新课标Ⅱ,18,12分)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.

解析(1)设数列{an}的首项为a1,公差为d,

∵bn=an?6,n为奇数,2an,n为偶数,∴b1=a1-6,b2=2a2,b3=a3-6,

又T3=16,且b1+b2+b3=a1+2a2+a3-12=4a2-12,∴4a2-12=16,∴a2=7,即a1+d=7,①

又S4=32,∴4a1+6d=32,②(2分)

由①②得a1=5,d=2,(每个结果1分)(4分)

∴an=5+2(n-1)=2n+3.(5分)

(2)证明:∵{an}为等差数列,

∴Sn=na1+n(n?1)2d=5n+n2-n=n2+4n,

①当n(n≥6)为偶数时,Tn=b1+b2+…+bn

=(b1+b3+…+bn-1)+(b2+b4+…+bn)

=(a1-6+a3-6+…+an-1-6)+(2a2+2a4+…+2an)

=a1+an?12·n2?6·n2+2·a2+an2·n2(7分)

=5+2(n?1)+32·n2?3n+7+2n+32·n=3n2+7n2.(9分)

∴Tn-Sn=3n2+7n2?n2?4n=3n2+7n?2n2?8n2=n2?n2=n(n?1)2>0(n≥6),

∴当n(n≥6)为偶数时,Tn>Sn.(10分)

②当n(n≥7)为奇数时,Tn=b1+b2+…+bn

=(b1+b3+…+bn)+(b2+b4+…+bn-1)

=(a1-6+a3-6+…+an-6)+(2a2+2a4+…+2an-1)

=a1+an2·n+12?6·n+12+2·a2+an?12·n?12=5+2n+32·n+12-3(n+1)+(n-1)·7+2(n?1)+32=3n2+5n?102,

∴Tn-Sn=3n2+5n?102-(n2+4n)=n2?3n?102=(n?5)(n+2)2>0(n≥7),

∴n(n≥7)为奇数时,Tn>Sn.(11分)

综上可知,当n>5时,Tn>Sn.(12分)

3.(2025届山东聊城期中教学质量检测,18)数列{an}中,若?d∈R,使得?n∈N*,都有an+2+2an+1+an=d成立,则称数列{an}为“三合定值数列”,已知a1=5,a2=-3,d=0.

(1)求a3,a4,a5;

(2)设bn=an+1+an,证明:数列{bn}为等比数列,并求an;

(3)设cn=(-2)nan,求数列{cn}的前n项和Sn.

解析(1)因为d=0,所以an+2+2an+1+an=0,又a1=5,a2=-3,

则a3+2a2+a1=0,即a3+2×(-3)+5=0,解得a3=1,

又a4+2a3+a2=0,即a4+2×1+(-3)=0,解得a4=1,

又a5+2a4+a3=0,即a5+2×1+1=0,解得a5=-3,

所以a3=1,a4=1,a5=-3.

(2)因为bn=an+1+an,则bn+1=an+2+an+1,

又an+2+2an+1+an=0,即an+2+an+1+an+1+an=0,所以bn+1+bn=0,

即bn+1=-bn,又b1=a2+a1=-3+5=2≠0,则bn+1bn=-1,

所以数列{bn}是以2为首项,以-1为公比的等比数列,

所以bn=2·(-1)n-1,

即an+1+an=2·(-1)n-1,

所以a2k+1+a2k=2·(-1)2k-1=-2①,k∈N*,

则a2k+a2k-1=2②,k∈N*,

两式相减可得a2k+1-a2k-1=-4,

即{an}的奇数项为等差数列,且a2k-1=5-4(k-1)=-4k+9,

令n=2k-1,则k=n+12,所以an=-2-2n+9=-2n+7,

又a2k+2+a2k+1=2·(-1)2k=2③,

由③-①可得a2k+2-a2k=4,k∈N*,

所以{an}的偶数项为等差数列,且a2k=-3+4(k-1)=4k-7,

令n=2k,则k=n2,即an=2n-7,

综上所述,an=?2n+7,n为奇数,2n?7,n为偶数.

(3)当n为奇数时,cn=-2n·(-2n+7)=2n·(2n-7),

当n为偶数时,cn=2n·(2n-7).

综上,cn=2n·(2n-7),

则Sn=2×(-5)+22×(-3)+23×(-1)+…+2n(2n-7),

2Sn=22×(-5)+23×(-3)+24×(-1)+…+2n(2n-9)+2n+1(2n-7),

两式相减可得-Sn=2×(-5)+2×22+2×23+…+2·2n-(2n-7)·2n+1,

-Sn=-10-6+2+22+23+…+2n+1-(2n-7)·2n+1,

-Sn=-16+2(1?2n+1)1?2-(2n-7)·2n+1,

-Sn=-16+2·2n+1-2-(2n-7)·2n+1,

-Sn=(-2n+9)·2n+1-18,所以Sn=(2n-9)·2n+1+18.

4.(2025届河北保定期中,18)已知数列{an}的首项a1=32,其前n项和Sn满足Sn=1-23an+1.

(1)求{an}的通项公式;

(2)设bn=nan,数列{bn}的前n项和为Tn,若Tn≤m对任意n∈N*恒成立,求m的最小值.

解析(1)当n=1时,由题意得a1=1-23a2,又∵a1=32,

∴a2=-34.(2分)

由Sn=1-23an+1①,可知Sn+1=1-23an+2②,

两式相减可得Sn+1-Sn=an+1=1?23an+2?1?23an+1=23an+1?23an+2,即an+2=-12an+1.(4分)

又∵a2=-12a1,(5分)

∴{an}是首项为32,公比为-12的等比数列,

故an=32·?12n?1.(6分)

(2)由(1)知bn=3n2·?12n?1.

Tn=3×12×?120+3×22×?121+3×32×?122+…+3(n?1)2·?12n?2+3n2·?12n?1①,

两边同乘-12,得-12Tn=3×12×?121+3×22×?122+3×32×?123+…+3(n?1)2·?12n?1+3n2·?12n②,

①-②得1+12Tn=32+32×?121+32×?122+32×?123+…+32·?12n?1?3n2·?12n,(8分)

从而32Tn=321??12n1??12?3n2·?12n=1??12n?3n2·?12n=1?1+3n2·?12n,(10分)

于是Tn=23?n+23?12n.(11分)

当n是奇数时,Tn=23+n+2312n,Tn+2-Tn=23+n+2+2312n+2?23+n+2312n=n+8312n+2?n+2312n=n+8312n+2?4n+23·12n+2=?3n·12n+20,Tn=23?n+2312n1)时,bn-1≥ak·bn;

(ii)求i=1Snbi.

解析(1)设等比数列{an}的公比为q,q>0,

则S2=a1+a2=a1+a1q=1+q=q2-1,∴q2-q-2=0,

∴q=2或q=-1(舍),∴an=2n-1,Sn=1?2n1?2=2n-1.

(2)(i)证明:当n=ak+1=2k时,bn=k+1.由题意,当k>1时,bak,bak+1,…,bn-1是一个以k为首项,2k为公差的等差数列,所以bn-1=k+(n-1-ak)×2k=(2k-1)k.因此bn-1-akbn=(2k-1)k-2k-1·(k+1)=2k-1(k-1)-k≥2(k-1)-k=k-2≥0.

所以,当n=ak+1(k∈N*,且k>1)时,bn-1≥akbn.

(i...