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2025新教材数学高考第一轮复习
7.5 数列综合
五年高考
考点 数列综合
1.(2022 新高考Ⅱ,17,10 分,中)已知{a }是等差数列,{b }是公比为 2 的等比数列,且 a -
n n 2
b =a -b =b -a .
2 3 3 4 4
(1)证明:a =b ;
1 1
(2)求集合{k|b =a +a ,1≤m≤500}中元素的个数.
k m 1
{a −6,n为奇数,
2.(2023新课标Ⅱ,18,12分,中)已知{a }为等差数列,b = n 记S ,T 分别为数
n n 2a ,n为偶数. n n
n
列{a },{b }的前n项和,S =32,T =16.
n n 4 3
(1)求{a }的通项公式;
n
(2)证明:当n>5时,T >S .
n n3.(2023天津,19,15分,难)已知{a }为等差数列,a +a =16,a -a =4.
n 2 5 5 3
2n−1
(1)求{a }的通项公式和∑ ❑a.
n i
i=2n−1
(2)设{b }为等比数列,当2k-1≤n≤2k-1(k∈N*)时,b 2时,a -b >0.
n n n
2.(2023 山东淄博一中校考 ,21)已知数列{a }是等差数列,{b }是等比数列,且
n n
b =2,b =4,a =b ,a +1=b .
2 3 1 1 8 5
(1)求数列{a }、{b }的通项公式;
n n
(2)设c
n
=a
n
+1,数列{c
n
}的前n项和为S
n
,若不等式(-1)nλ√b
2 b b b b 2n+1
1 3 5 2n−17.5 数列综合
五年高考
考点 数列综合
1.(2022 新高考Ⅱ,17,10 分,中)已知{a }是等差数列,{b }是公比为 2 的等比数列,且 a -
n n 2
b =a -b =b -a .
2 3 3 4 4
(1)证明:a =b ;
1 1
(2)求集合{k|b =a +a ,1≤m≤500}中元素的个数.
k m 1
解析 (1)证明:设等差数列{a }的公差为d.
n
由a -b =a -b 得a +d-2b =a +2d-4b ,故d=2b ,①
2 2 3 3 1 1 1 1 1
由a -b =b -a 得a +2d-4b =8b -a -3d,
3 3 4 4 1 1 1 1
故2a +5d=12b ,②
1 1
由①②得2a +10b =12b ,即a =b .
1 1 1 1 1
(2)由(1)知d=2b =2a ,由b =a +a ,1≤m≤500得b ×2k-1=2a +(m-1)d,即a ×2k-1=2a +2(m-1)a ,其
1 1 k m 1 1 1 1 1 1
中a ≠0,∴2k-1=2m,即2k-2=m,∴1≤2k-2≤500,∴0≤k-2≤8,
1
∴2≤k≤10.
故集合{k|b =a +a ,1≤m≤500}中元素的个数为9.
k m 1
2.(2023新课标Ⅱ,18,12分,中)已知{a }为等差数列,b ={a −6,n为奇数,记S ,T 分别为数
n n n n n
2a ,n为偶数.
n
列{a },{b }的前n项和,S =32,T =16.
n n 4 3
(1)求{a }的通项公式;
n
(2)证明:当n>5时,T >S .
n n解析 (1)设数列{a }的首项为a ,公差为d,
n 1
∵b ={a −6,n为奇数, ∴b =a -6,b =2a ,b =a -6,
n n 1 1 2 2 3 3
2a ,n为偶数,
n
又T =16,且b +b +b =a +2a +a -12=4a -12,∴4a -12=16,∴a =7,即a +d=7,①
3 1 2 3 1 2 3 2 2 2 1
又S =32,∴4a +6d=32,② (2分)
4 1
由①②得a =5,d=2,(每个结果1分) (4分)
1
∴a =5+2(n-1)=2n+3. (5分)
n
(2)证明:∵{a }为等差数列,
n
n(n−1)
∴S =na + d=5n+n2-n=n2+4n,
n 1
2
①当n(n≥6)为偶数时,T =b +b +…+b
n 1 2 n
=(b +b +…+b )+(b +b +…+b )
1 3 n-1 2 4 n
=(a -6+a -6+…+a -6)+(2a +2a +…+2a )
1 3 n-1 2 4 n
a +a n n a +a n
= 1 n−1· −6· +2· 2 n· (7分)
2 2 2 2 2
5+2(n−1)+3 n 7+2n+3 3n2+7n
= · −3n+ ·n= . (9分)
2 2 2 2
3n2+7n 3n2+7n−2n2−8n n2−n n(n−1)
∴T -S = −n2−4n= = = >0(n≥6),
n n
2 2 2 2
∴当n(n≥6)为偶数时,T >S . (10分)
n n
②当n(n≥7)为奇数时,T =b +b +…+b
n 1 2 n
=(b +b +…+b )+(b +b +…+b )
1 3 n 2 4 n-1
=(a -6+a -6+…+a -6)+(2a +2a +…+2a )
1 3 n 2 4 n-1
a +a n+1 n+1 a +a n−1 5+2n+3 n+1
= 1 n· −6· +2· 2 n−1· = · -3(n+1)+(n-1)·
2 2 2 2 2 2 2
7+2(n−1)+3 3n2+5n−10
= ,
2 2
3n2+5n−10 n2−3n−10 (n−5)(n+2)
∴T -S = -(n2+4n)= = >0(n≥7),
n n
2 2 2
∴n(n≥7)为奇数时,T >S . (11分)
n n
综上可知,当n>5时,T >S . (12分)
n n
3.(2023天津,19,15分,难)已知{a }为等差数列,a +a =16,a -a =4.
n 2 5 5 3
(1)求{a }的通项公式和 2n−1a.
n ∑ ❑ i
i=2n−1(2)设{b }为等比数列,当2k-1≤n≤2k-1(k∈N*)时,b (a ) =2k+1-1,即b >2k-1,
k n min k+1 n max k
∴2k-10.
n 1
因为c ≤b ≤c ,所以qk-1≤k≤qk,其中k=1,2,3,…,m.
k k k+1
当k=1时,有q≥1;
lnk lnk
当k=2,3,…,m时,有 ≤ln q≤ . (9分)
k k−1
lnx 1−lnx
设f(x)= (x>1),则f '(x)= .
x x2
令f '(x)=0,得x=e.列表如下:
x (1,e) e (e,+∞)
f '(x) + 0 -
极大
f(x) ↗ ↘
值
ln2 ln8 ln9 ln3 ln3
因为 = < = ,所以f(k) =f(3)= . (11分)
max
2 6 6 3 3lnk
取q=√33,当k=1,2,3,4,5时, ≤ln q,即k≤qk,经检验知qk-1≤k也成立.
k
因此所求m的最大值不小于5. (14分)
若m≥6,分别取k=3,6,得3≤q3,且q5≤6,从而q15≥243,且q15≤216,所以q不存在.
因此所求m的最大值小于6.
综上,所求m的最大值为5. (16分)
三年模拟
综合拔高练1
1.(2023山东潍坊二模,18)已知等差数列{a }的公差为2,前n项和为S ,且S ,S ,S 成等比数
n n 1 2 4
列.
(1)求数列{a }的通项公式;
n
(2)若b
n
= a
n
,n∈N*,求数列{b
n
}的最大项.
a a +6
n n+1
解析 (1)由题意知S =na +n(n-1),
n 1
又因为 =S ·S ,即(2a +2)2=a ·(4a +12),所以a =1,
S2 1 4 1 1 1 1
2
又d=2,所以a =2n-1.
n
2n−1 2n−1
(2)由(1)知b = = ,
n (2n−1)(2n+1)+6 4n2+5
t+1
设t=2n-1(t=1,3,5,…),所以n= ,又因为t>0,
2
t t 1
= =
所以f(t)= (t+1) 2+5 t2+2t+6 6 (t=1,3,5,…),
t+ +2
t
因为函数在t≥3时递减,所以f(t)的最大值可能出现在t=1或t=3时,
1 1
t=1时,n=1,b = = ,
1 1+6+2 9
1 1 1
t=3时,n=2,b = = > ,
2 3+2+2 7 9
1
所以数列{b }的最大项为b = .
n 2
7
2.(2023浙江宁波二模,20)已知等比数列{a }的前n项和S 满足a =S +1(n∈N*).
n n n+1 n
(1)求首项a 的值及{a }的通项公式;
1 na
(2)设b
n
=log
2
2n+1 (n∈N*),求满足a
n
-b
n
<2 023的最大正整数n的值.
a ·a ·a ·…·a
2 22 23 2n
解析 (1)当n≥2时,由a =S +1得a =S +1,则a -a =S -S ,即a =2a (n≥2),
n+1 n n n-1 n+1 n n n-1 n+1 n
因为数列{a }是等比数列,所以公比为2,
n
当n=1时,a =a +1,即2a =a +1,所以a =1,
2 1 1 1 1
所以{a }的通项公式为a =2n-1.
n n
(2)由(1)得 a =22n−1 ,所以log 2a =2n-1,
2n 2n
故b =log -(log a +log )
n 2a 2 2 2a +…+log2a
2n+1 22 2n
=2n+1-1-(21-1+22-1+…+2n-1)
2(1−2n
)
=2n+1+n-1- =n+1,
1−2
由a -b <2 023得2n-1-(n+1)<2 023,即2n-1-n<2 024,
n n
令f(n)=2n-1-n,
则f(n+1)-f(n)=2n-(n+1)-(2n-1-n)=2n-1-1,
所以f(n)在n≥2时递增,且f(2)=f(1)=0,
而f(11)=210-11=1 013<2 024, f(12)=211-12=2 036>2 024,
所以满足条件的最大正整数n=11.
3.(2024届云南师大附中适应性月考,19)各项均为正数的等比数列{a }中,a =1,a =4a .
n 1 5 3
(1)求{a }的通项公式;
n
1 2 n
(2)设m为整数,且对任意的n∈N*,m≥ + +…+ ,求m的最小值.
a a a
1 2 n
解析 (1)设公比为q,由a =4a ,得a q2=4a ,得q2=4,
5 3 3 3
又a >0,所以q>0,所以q=2,则a =a qn-1=2n-1.
n n 1
n n
(2)由(1)知,a =2n-1,故 = ,
n a 2n−1
n
1 2 3 n 2 3 n
设S = + + +…+ =1+ + +…+ ,
n a a a a 21 22 2n−1
1 2 3 n
1 1 2 3 n
则 Sn= + + +…+ ,
2 2 22 23 2n
1 1 1 1 1 n
所以S - Sn=1+ + + +…+ − ,
n 2 2 22 23 2n−1 2n1
1−
1 2n n 2 n n+2
所以 Sn= − =2− − =2− ,
2 1 2n 2n 2n 2n
1−
2
n+2
所以S =4- .
n 2n−1
n+2
当n趋近于正无穷时, 趋近于0,所以S 趋近于4且S <4,所以m≥4,所以m的最小值为
2n−1 n n
4.
4.(2024届河北邯郸一中月考,18)已知数列{a }满足4S -2a =2n,n∈N*,其中S 为{a }的前n
n n n n n
项和.证明:
(1){a
n−
1}是等比数列;
2n 6
(2) 1 1 1 1 <1.
+ + +…+
6a −3 6a +3 6a −3 6a +3×(−1) n
1 2 3 n
证明 (1)∵4S -2a =2n,∴4S -2a =2n-1(n≥2),
n n n-1 n-1
两式相减得4(S -S )-2a +2a =2n-2n-1,
n n-1 n n-1
即a +a =2n-2(n≥2).
n n-1
a 1 −a +2n−2 1 1( a 1)
n− n−1 − − n−1−
∴ 2n 6 2n 6 2 2n−1 6 1(n≥2).
= = =−
a 1 a 1 a 1 2
n−1− n−1− n−1−
2n−1 6 2n−1 6 2n−1 6
当n=1时,4S -2a =21,即a =1.
1 1 1
又∵ a
1−
1
=
1≠0,∴{a
n−
1}
是以
1为首项,-1为公比的等比数列.
21 6 3 2n 6 3 2
(2)由(1)得a n− 1 = 1 × ( − 1) n−1,
2n 6 3 2
2 1
所以a =- ×(-1)n+ ×2n.
n
3 6
令b = 1 1 ,
n =
6a +3(−1) n 2n−(−1) n
n3 3
×22n ×22n
则b +b = 1 1 2 2 3 .
2n-1 2n + = < =
22n−1+1 22n−1 24n−1+22n−1−1 24n−1 22n
3 ( 1 )
× 1−
数列{b }的前2n项和T <3 3 3 4 4n <1.
n 2n + +…+ =
4 42 4n 1
1−
4
又b >0,∴T 2时,a -b >0.
n n n
解析 (1)因为a =2S +2,所以n≥2时,a =2S +2,
n+1 n n n-1
所以a -a =2(S -S )=2a ,所以a =3a (n≥2),
n+1 n n n-1 n n+1 n
因为{a }为等比数列,所以a =3a ,
n 2 1
又因为a =2S +2=2a +2,所以a =2,
2 1 1 1
则等比数列{a }的首项为2,公比为3,所以a =2·3n-1.
n n
(2)证明:要证n>2时,a -b >0,即证n>2时,2·3n-1>2n+1,
n n
需证n>2时,(3) n−1>2,
2
因为n>2且n∈N*,所以n≥3,n-1≥2,
又3>1,所以 (3) n−1 9>2,原不等式成立.
≥
2 2 4
2.(2023 山东淄博一中校考 ,21)已知数列{a }是等差数列,{b }是等比数列,且
n n
b =2,b =4,a =b ,a +1=b .
2 3 1 1 8 5
(1)求数列{a }、{b }的通项公式;
n n
(2)设c
n
=a
n
+1,数列{c
n
}的前n项和为S
n
,若不等式(-1)nλ-2.
2n−2
综上可得,实数λ的取值范围是(-2,3).
3.(2024届湖北黄冈浠水一中校考,20)已知数列{a }的各项均为正数,其前n项和S 满足2
n n
1
√S =a +1,数列{b }满足b = .
n n n n (a +1)(a +1)
n n+1
(1)求{a }的通项公式;
n
5m−2
(2)设数列{b }的前n项和为T ,若 0得a -a =2,
n n-1 n n-1 n n n-1
∴数列{a }是以1为首项,2为公差的等差数列,
n
故a =1+2(n-1)=2n-1.
n
(2)由(1)可得b = 1 1 1(1 1 ),
n = = −
(a +1)(a +1) 2n×2(n+1) 4 n n+1
n n+1
∴数列{b n }的前n项和T n =1 1− 1 + 1 − 1 +…+ 1 − 1 = 1( 1− 1 ) = n .
4 2 2 3 n n+1 4 n+1 4(n+1)
n+1 n 1
∵T -T = − = >0,
n+1 n 4(n+2) 4(n+1) 4(n+1)(n+2)
1
∴T 递增,∴T ≥T = ,
n n 1
8
n 1 1 1
∵T = < ,∴ ≤Tn< ,
n 4(n+1) 4 8 4
1
{ ≤5m,
若5m−2√b
2 b b b b 2n+1
1 3 5 2n−1
解析 (1)3a =2(S +2n),当n=1时,3a =2(S +2),解得a =4,
n n 1 1 1
当n≥2时,3a =2(S +2n-2),两式相减得a =3a +4,
n-1 n-1 n n-1
所以a +2=3(a +2),n≥2,
n n-1所以{a +2}是首项为6,公比为3的等比数列,
n
即a +2=6×3n-1,所以a =6×3n-1-2.
n n
a +2 6×3n−1
(2)证明:由(1)可得b =log n =log3 =log 3n=n,
n 3 3
2 2
即证(1+1)( 1)( 1) ( 1 ) .
1+ 1+ … 1+ >√2n+1
3 5 2n−1
( 1 )( 1 )( 1 ) ( 1 )
证法一:令f(n)= 1+ 1+ 1+ … 1+ ,
b b b b
1 3 5 2n−1
√2n+1
f(n+1) 2n+2
则 = ,
f(n) √2n+1·√2n+3
因为(2n+2)2>(2n+1)(2n+3),所以f(n+1)>f(n),
2
所以f(n)递增,即f(n)>f(1)= >1,
√3
即( 1 )( 1 )( 1 ) ( 1 ) .
1+ 1+ 1+ … 1+ >√b
b b b b 2n+1
1 3 5 2n−1
证法二(放缩法): 2n 2n+1 ( 2n ) 2 2n 2n+1 2n+1,
> ⇒ > · =
2n−1 2n 2n−1 2n−1 2n 2n−1
所以(2) 2 3,(4) 2 5,……,( 2n ) 2 2n+1,
> > >
1 1 3 3 2n−1 2n−1
相 乘 得 (2) 2 (4) 2 ( 2n ) 2 3 5 2n+1=2n+1, 即
· ·…· > · ·…·
1 3 2n−1 1 3 2n−1
( 1 )( 1 )( 1 ) ( 1 ) .
1+ 1+ 1+ … 1+ >√b
b b b b 2n+1
1 3 5 2n−1
