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微专题18 数列中的放缩问题
一．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
.
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
.
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.
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
.
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.
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).
【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.
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.
