微专题18 数列中的放缩问题
近几年高考：
1.数列放缩是高考重点考查的内容之一,数列与不等式结合试
题将稳定在中等偏难的程度,其核心技能是放缩技巧的应用.
2.通项放缩常见结论
(1) 1
n
2>
1
n(n+1)
=1
n
- 1
n+1
;
(2) 1
n
2<
1
n
2−n
= 1
n−1
-1
n
(n≥2);
(3) 1
n
2<
1
n
2−1
=1
2(
1
n−1 −
1
n+1)(n≥2);
(4) 1
n
2= 4
4 n
2 <
4
4 n
2−1
=2(
1
2n−1 −
1
2n+1);
(5)Tr+1=Cn
r· 1
n
r =
n!
r !(n−r )!
· 1
n
r < 1
r !
<
1
r(r −1)
= 1
r −1
-1
r
(r≥2);
(6)(1+ 1
n)
n
<1+1+ 1
1×2
+ 1
2×3
+…+
1
(n−1)n
<3;
(7) 1
❑√n
=
2
❑√n+ ❑√n
<
2
❑√n−1+ ❑√n
=2(-❑√n−1+❑√n)(n≥2);
(8) 1
❑√n
=
2
❑√n+ ❑√n
>
2
❑√n+ ❑√n+1
=2(-❑√n+❑√n+1);
(9) 1
❑√n
=
2
❑√n+ ❑√n
<
2
❑√n−1
2 + ❑√n+ 1
2
=
2❑√2
❑√2n−1+ ❑√2n+1
=❑√2(-❑√2n−1+❑√2n+1);
(10)
2
n
(2
n−1)
2 =
2
n
(2
n−1)(2
n−1)
<
2
n
(2
n−1)(2
n−2)
=
2
n−1
(2
n−1)(2
n−1−1)
=
1
2
n−1−1
-
1
2
n−1
(n≥2);
(11) 1
❑√n
3 =
1
❑√n·n
2 <
1
❑√(n−1)n(n+1)
=
❑√n+1−❑√n−1
❑√(n−1)n(n+1)
·
1
❑√n+1−❑√n−1
=[
1
❑√(n−1)n −
1
❑√n(n+1)]·
1
❑√n+1−❑√n−1
=(
1
❑√n−1 −
1
❑√n+1)·
❑√n+1+ ❑√n−1
2❑√n
<2(
1
❑√n−1 −
1
❑√n+1)(n≥2);
(12) 1
❑√n
3 =
2
❑√n
2·n+
❑√n·n
2<
2
n ❑√n−1+(n−1)❑√n
=
2
❑√(n−1)n(❑√n+ ❑√n−1)
=
2(❑√n−❑√n−1)
❑√(n−1)n
=
2
❑√n−1
- 2
❑√n
(n≥2);
(13)
1
2
n−1
=
1
(1+1)
n−1
<
1
Cn
0+Cn
1+Cn
2−1
=
2
n(n+1)
=2
n
- 2
n+1
;
(14)
1
2
n−1
<
2
n−1
(2
n−1−1)(2
n−1)
=
1
2
n−1−1
-
1
2
n−1
(n≥2).
二．典型例题
1. 先求和再放缩证明不等式
例1 (2025·成都诊断)已知数列{an}满足a1+2a2+…
+nan=(n-1)·2n+1+2.
(1)求{an}的通项公式;
(2)设bn= 1
an
+ 1
an
2 ,证明:b1+b2+…+bn<4
3
.
(1)解 由题意可知,当n=1 时,a1=2;
当n≥2 时,由a1+2a2+…+nan=(n-1)·2n+1+2 得,
a1+2a2+…+(n-1)an-1=(n-2)·2n+2,
两式作差可得,nan=(n-1)·2n+1-(n-2)·2n=n·2n,∴an=2n,
a1=2 也适合该式,故an=2n.
(2)证明 由题意知bn= 1
an
+ 1
an
2 = 1
2
n + 1
4
n ,
故b1+b2+…+bn=
1
2(1−1
2
n)
1−1
2
+
1
4(1−1
4
n)
1−1
4
=1- 1
2
n +1
3
-1
3
×
1
4
n =4
3
-(
1
2
n +
1
3×4
n),
由于n∈N*,则1
2
n +
1
3×4
n >0,
故4
3
-(
1
2
n +
1
3×4
n)<4
3
,
即b1+b2+…+bn<4
3
.
规律方法 此类不等式一般另一端为常数,求和以后常利用去
项放缩或利用函数的单调性放缩.
训练1 (2025·武汉模拟)已知数列{an}的首项为2,前n 项和为
Sn,且Sn+1+2=an+3n+Sn.
(1)求数列{an}的通项公式;
(2)求满足an>3n+5
2
的n 的最小值;
(3)已知bn=
1
6an+6n−20
,记数列{bn}的前n 项和为Tn,求
证:-1
2
≤Tn<-1
3
.
(1)解 由已知Sn+1+2=an+3n+Sn,
则an+1=Sn+1-Sn=an+3n-2,
即an+1-an=3n-2,
则an-an-1=3n-5,an-1-an-2=3n-8,…,a2-a1=1,
等式左右分别相加可得an-a1=(3n-5)+(3n-8)+…+1
=(3n−5+1)(n−1)
2
=3n
2−7n+ 4
2
,
则an=3n
2−7n+ 4
2
+a1=3n
2−7n+8
2
.
(2)解 由(1)得an=3n
2−7n+8
2
,
且an>3n+5
2
,即3n
2−7n+8
2
>3n+5
2
,
化简可得(3n-1)(n-3)>0,
又n∈N*,即n>3,
所以满足an>3n+5
2
的n 的最小值为4.
(3)证明 依题意得bn=
1
6an+6n−20
=
1
9n
2−15n+ 4
=
1
(3n−4)(3n−1)
=1
3(
1
3n−4 −
1
3n−1),
则Tn=
1
3(−1−1
2 + 1
2 −1
5 + 1
5 −1
8 +…+
1
3n−4 −
1
3n−1)
=1
3(−1−
1
3n−1)=-1
3
-
1
9n−3
,
又n∈N*,所以
1
9n−3
∈(0, 1
6],
所以Tn=-1
3
-
1
9n−3
∈[−1
2 ,−1
3),
即-1
2
≤Tn<-1
3
.
2. 先放缩通项再求和证明不等式
例2 设数列{an}的前n 项和为Sn,a1=1,2Sn=n2+n(n∈N*).
(1)求数列{an}的通项公式;
(2)设数列{bn}的前n 项和为Tn,且bn=
1
❑√an+ ❑√an+1
,求T99;
(3)求证:
1
2❑√a1
+
1
2❑√a2
+
1
2❑√a3
+…+
1
2❑√a99
>9.
(1)解 因为2Sn=n2+n,①
当n≥2 时,2Sn-1=(n-1)2+n-1,②
所以①-②得到2an=n2+n-(n-1)2-(n-1)=2n,即an=n,
又a1=1,满足an=n,所以an=n.
(2)解 因为bn=
1
❑√an+ ❑√an+1
=
1
❑√n+ ❑√n+1
=❑√n+1-❑√n,
所以T99=b1+b2+…+b99=❑√2-1+❑√3-❑√2+…+❑√100-❑√99=
❑√100-1=9.
(3)证明 因为
1
2❑√an
= 1
2❑√n
>
1
❑√n+1+ ❑√n
=❑√n+1-❑√n,
所以
1
2❑√a1
+
1
2❑√a2
+…+
1
2❑√a99
= 1
2❑√1
+ 1
2❑√2
+…+
1
2❑√99
>❑√2-1+❑√3-❑√2+…+❑√100-❑√99=❑√100-1=9,
即
1
2❑√a1
+
1
2❑√a2
+
1
2❑√a3
+…+
1
2❑√a99
>9.
规律方法 此类题型关键是如何放缩数列的通项,需要熟悉常
见的放缩技巧及结论.
训练2 (2025·巴中模拟)已知数列{an}的通项公式为an=2
n−1
2
n
.
(1)求证:1
2
≤an<1;
(2)令bn=log2
1
1−an
,求证: 1
b1
2 + 1
b2
2+…+ 1
bn
2 <7
4
.
证明 (1)an=2
n−1
2
n
=1- 1
2
n ,
可知数列{an}单调递增,
则an=1- 1
2
n <1,
则当n=1 时,{an}取最小值为1
2
,
故1
2
≤an<1,得证.
(2)bn=log2
1
1−an
=n,
当n≥3 时, 1
bn
2 = 1
n
2<
1
n(n−1)
= 1
n−1
-1
n
,
1
b1
2 + 1
b2
2+…+ 1
bn
2 <1+1
4
+1
2
-1
3
+…+ 1
n−1
-1
n
=7
4
-1
n
<7
4
,得证.
3. 通项放缩与求值
例3 已知各项均为正数的数列{an}的前n 项和为Sn,且an,Sn,an
2
为等差数列.
(1)求数列{an}的通项公式;
(2)若m 为正整数,记集合{an|
an
2 + 2
an
≤m}的元素个数为
bm,求数列{bn}的前50 项和T50.
解 (1)由已知得2Sn=an+an
2,
当n=1 时,有a1=1,
当n≥2 时,有2Sn-1=an-1+an−1
2
,
∴2an=an
2-an−1
2
+an-an-1,
(an-an-1)(an+an-1)=an+an-1,
an+an-1>0,
∴an-an-1=1,
∴数列{an}为等差数列,∴an=n.
(2)由an
2
+ 2
an
≤m,得m-❑√m
2−4≤an≤m+❑√m
2−4,
令m=1,m=2,显然b1=0,b2=1.
当m≥3 时,令m-❑√m
2−4≤1,
得(m-1)2<m2-4,得m>5
2
,
易证2m-1<m+❑√m
2−4<2m,
∴1≤an≤2m-1,bm=2m-1,
从而{bn}的前50 项和
T50=0+1+(5+7+…+99)=1+48×104
2
=2 497.
规律方法 1.通项放缩确定新数列注意解相关不等式;
2.先放缩再求和式及和式的应用,应注意考虑所得式子的性质.
训练3 已知Tn为正项数列{an}的前n 项的乘积,且a1=3,T n
2=
an
n+1.
(1)求数列{an}的通项公式;
(2)设bn=an−1
an+1
,数列{bn}的前n 项和为Sn,求[S2 026]([x]表示不
超过x 的最大整数).
解 (1)当n≥2 时,T n
2=an
n+1,T n−1
2
=an−1
n
,
相除得an
2= an
n+1
an−1
n
,an
n−1=an−1
n
,
an
1
n=an−1
1
n−1,n≥2,
∴数列{an
1
n}是常数列,a1=3,
∴an
1
n=3,an=3n.
(2)bn=3
n−1
3
n+1
=1-
2
3
n+1
>1- 2
3
n ,
∴Sn=(1−
2
3
1+1)+(1−
2
3
2+1)+…+(1−
2
3
n+1)=n-
(
2
3
1+1
+…+
2
3
n+1)<n,
而Sn>n-(
2
3
1 + 2
3
2 +…+ 2
3
n)=n-
2
3[1−(
1
3)
n
]
1−1
3
=n-1+ 1
3
n
>n-1,
∴2 025<S2 026<2 026,故[S2 026]=2 025.
【精准强化练】
1.(2025·重庆部分学校联考)已知数列{an}满足a1=1,an+an+1=2n+1.
(1)证明:数列{
an
2
n −2
3}为等比数列,并求数列{an}的通项公式;
(2)设Tn= 1
a1
+ 1
a2
+…+ 1
an
,证明:Tn<7
4
对任意n∈N*恒成立.
证明 (1)由an+an+1=2n+1,
得an+1
2
n+1 +1
2
×an
2
n =1,
因为an+1
2
n+1 -2
3
=-1
2(
an
2
n −2
3),
a1
2
1 -2
3
=-1
6
,
所以数列{
an
2
n −2
3}是首项为-1
6
,
公比为-1
2
的等比数列,
因此,an
2
n -2
3
=-1
6
×(−1
2)
n−1
,
所以an=2
n+1+(−1)
n
3
.
(2)由已知得a1=1,a2=3,a3=5,…,
显然Tn单调递增,T1<T2= 1
a1
+ 1
a2
=4
3
<7
4
.
当n>2 且n 是奇数时,
1
an−1
+ 1
an
=3(
1
2
n+1
+
1
2
n+1−1)
=3×
2
n+2
n+1
(2
n+1)(2
n+1−1)
<3× 2
n+2
n+1
2
n×2
n+1
=3(
1
2
n + 1
2
n+1),
所以Tn= 1
a1
+ 1
a2
+…+ 1
an
<1+3×[(
1
2
3 + 1
2
4)+…+(
1
2
n + 1
2
n+1)]
=1+3×(
1
4 −1
2
n+1)=7
4
- 3
2
n+1<7
4
.
当n>2 且n 是偶数时,n+1 是奇数,有Tn<Tn+1<7
4
,
所以对任意n∈N*,
Tn= 1
a1
+ 1
a2
+…+ 1
an
<7
4
.
2.(2025·太原调研)已知数列{an}满足a1=1,a2n+1=a2n+1,a2n=2a2n-1.
(1)求数列{an}的通项公式;
(2)设Tn= 1
a1
+ 1
a2
+…+ 1
an
,求证:T2n<3.
(1)解 将a2n=2a2n-1代入a2n+1=a2n+1 中,得a2n+1=2a2n-1+1.
下面构造等比数列:
令a2n+1-k=2(a2n-1-k),
得a2n+1=2a2n-1-k,则-k=1,则k=-1,
∴a2n+1+1=2(a2n-1+1),
故数列{a2n-1+1}是首项为2,公比为2 的等比数列,
∴a2n-1+1=2·2n-1,∴a2n-1=2n-1,
a2n=2·a2n-1=2(2n-1)=2n+1-2,
故an={
2
n+1
2 −1,n 为奇数,
2
n+2
2 −2,n 为偶数.
(2)证明 由(1)知a2n-1=2n-1,a2n=2n+1-2,
∴
1
a2n−1
=
1
2
n−1
, 1
a2n
=
1
2
n+1−2
.
又2n-1≥2n-1,∴
1
2
n−1
≤1
2
n−1,
2n+1-2≥2n,∴
1
2
n+1−2
≤1
2
n ,
故
1
a2n−1
≤1
2
n−1, 1
a2n
≤1
2
n ,
则T2n= 1
a1
+ 1
a2
+…+
1
a2n−1
+ 1
a2n
<[(
1
2)
0
+(
1
2)
1
+(
1
2)
2
+…+(
1
2)
n−1
]+
[(
1
2)
1
+(
1
2)
2
+…+(
1
2)
n
]
=
1×[1−(
1
2)
n
]
1−1
2
+
1
2 ×[1−(
1
2)
n
]
1−1
2
=2[1−(
1
2)
n
]+[1−(
1
2)
n
]=3-3(
1
2)
n
<3,
故T2n<3.
