数列大题专项练习
1.[2018·全国卷Ⅱ]记Sn 为等差数列{an }的前n 项和,已知a 1=-7,S 3=-15. (1)求{an }的通项公式; (2)求Sn ,并求Sn 的最小值.
解析:(1)解:设{an }的公差为d ,由题意得3a 1+3d =-15.由a 1=-7得d =2. 所以{an }的通项公式为an =a 1+(n -1)d =2n -9. (2)解:由(1)得Sn =
a 1+an
2
·n =n 2-8n =(n -4)2
-16.
所以当n =4时,Sn 取得最小值,最小值为-16.
2.[2019·河北廊坊省级示范高中联考]在数列{a n }中,a 1=1,a n +1a n =4(n +1)2
n (n +2),设b n =
n +1
n
·a n . (1)证明:数列{b n }是等比数列; (2)求{a n }的前n 项积T n .
解析:(1)因为b n +1b n =n +2
n +1·a n +1
n +1n
·a n =n (n +2)(n +1)2·
a n +1a n =n (n +2)(n +1)2·4(n +1)2
n (n +2)
=4,b 1=2a 1=2, 所以数列{b n }是首项为2,公比为4的等比数列. (2)由(1)知b n =
n +1n ·a n =2·4n -1,则a n =n n +1
·22n -1
. 从而T n =? ??
??12×23×3
4×…×n n +1·21+3+5+…+(2n -1)
=2n
2
n +1
. 3.[2019·辽宁鞍山月考]已知数列{a n }的前n 项和为S n ,a 1+a 2=4,2S n +1-a n +1=2S n +3a n (n ∈N *).
(1)求数列{a n }的通项公式;
(2)设b n =3n
(a n +1-1)S n +1,数列{b n }的前n 项和为T n ,证明:38≤T n <1
2.
解析:(1)∵2S n +1-a n +1=2S n +3a n ,∴2a n +1-a n +1=3a n , ∴a n +1=3a n (n ∈N *
),∵a 1+a 2=4,∴a 1=1, ∴数列{a n }是首项为1,公比为3的等比数列, ∴a n =3
n -1
.
(2)由(1)知S n =3n
-1
2
.
∵b n =3n
(a n +1-1)S n +1,∴b n =2×3n
(3n -1)(3n +1
-1)=13n -1-1
3n +1-1, ∴T n =?
????131-1-132-1+? ????132-1-133-1+…+? ???
?13n -1-13n +1-1=12-13n +1-1
.
∵n ∈N *
,所以-13n +1-1∈??????-18,0,
∴38≤12-13n +1-1<12,即38≤T n <12
. 4.[2019·湖南衡阳联考]已知数列{a n },{b n }满足a 1=1,b 1=12,2a n +1=a n +1
2
b n ,2b n +
1
=12
a n +
b n (n ∈N *
). (1)证明:数列{a n +b n },{a n -b n }均是等比数列; (2)记S n 为数列{a n }的前n 项和,S n =λ+μ? ????a n -89×14n , 求λ-μ的值.
解析:(1)依题意得???
??
2a
n +1
=a n +1
2
b n ,
2b
n +1=1
2
a n +
b n ,
两式相加,
得a n +1+b n +1=3
4
(a n +b n ),∴{a n +b n }为等比数列;
两式相减,得a n +1-b n +1=1
4(a n -b n ),∴{a n -b n }为等比数列.
(2)∵a 1=1,b 1=12,∴a 1+b 1=32,a 1-b 1=1
2.
由(1)可得a n +b n =32×? ??
??34n -1
①,
a n -
b n =12
×? ??
??14
n -1 ②.
①+②,得 a n =? ????14n +? ??
??34n
,
∴S n =14×? ?
???1-14n 1-14+34×? ????1-3n
4n 1-34
=13×? ????1-14n +3×? ????1-3n
4n =103-13×14n -3×? ????34n
.
又S n =λ+μ? ????a n -89×14n =λ+μ??????19×14n +? ????34n ,∴λ=103,μ=-3,∴λ-μ=193. 5.[2019·河南洛阳孟津二中月考]在数列{a n }中,设f (n )=a n ,且f (n )满足f (n +1)-2f (n )=2n
(n ∈N *
),a 1=1.
(1)设b n =a n
2n -1,证明:数列{b n }为等差数列;
(2)求数列{3a n -1}的前n 项和S n . 解析:(1)由已知得a n +1=2a n +2n
,得b n +1=
a n +12
n
=
2a n +2n
2
n
=a n
2
n -1+1=b n +1, ∴b n +1-b n =1,又a 1=1,∴b 1=1,∴{b n }是首项为1,公差为1的等差数列. (2)由(1)知,b n =a n
2
n -1=n ,∴a n =n ·2
n -1,
3a n -1=3n ·2
n -1
-1.
∴S n =3×1×20
+3×2×21
+3×3×22
+…+3(n -1)×2
n -2
+3n ×2n -1
-n ,
两边同时乘以2,得2S n =3×1×21
+3×2×22
+…+3(n -1)×2n -1
+3n ×2n
-2n ,
两式相减,得-S n =3×(1+21
+22
+…+2n -1
-n ×2n )+n =3×(2n
-1-n ×2n
)+n =3(1
-n )2n
-3+n ,
∴S n =3(n -1)2n
+3-n .
6.[2019·河北九校第二次联考]已知数列{a n }是各项都为正数的数列,其前n 项和为
S n ,且S n 为a n 与1
a n
的等差中项.
(1)求数列{a n }的通项公式;
(2)设b n =(-1)
n
a n
,求{b n }的前n 项和T n .
解析:(1)由题意知2S n =a n +1a n
,即2S n a n -a 2
n =1,(※)
当n =1时,由(※)式可得S 1=1;
当n ≥2时,a n =S n -S n -1,代入(※)式,得2S n (S n -S n -1)-(S n -S n -1)2
=1, 整理得S 2
n -S 2
n -1=1.
所以{S 2
n }是首项为1,公差为1的等差数列,所以S 2
n =1+(n -1)×1=n . 因为数列{a n }的各项都为正数,所以S n =n . 由此可得a n =S n -S n -1=n -n -1(n ≥2), 又a 1=S 1=1,所以a n =n -n -1. (2)由(1)知b n =(-1)
n
a n
=
(-1)
n
n -n -1
=(-1)n
(n +n -1).
当n 为奇数时,
T n =-1+(2+1)-(3+2)+…+(n -1+n -2)-(n +n -1)=-n ;
当n为偶数时,
T n=-1+(2+1)-(3+2)+…-(n-1+n-2)+(n+n-1)=n. 所以{b n}的前n项和T n=(-1)n n.