2020高考文科数学二轮提分广西等课标3卷专用题型练：4 大题专项2
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题型练4 大题专项(二)数列的通项、求和问题
1.(2019湖北4月调研,17)已知数列{a n }满足a 2-a 1=1,其前n 项和为S n ,当n ≥2时,S n-1-1,S n ,S n+1成等差数列.
(1)求证{a n }为等差数列;(2)若S n =0,S n+1=4,求n.
2.(2019吉林实验中学检测,17)已知等差数列{a n }满足a 4=7,2a 3+a 5=19.(1)求a n ;
(2)设{b n -a n }是首项为2,公比为2的等比数列,求数列{b n }的通项公式及前n 项和T n .
3.设{a n }是等差数列,且a 1=ln 2,a 2+a 3=5ln 2.(1)求{a n }的通项公式.(2)求+…+.
e
a 1
+e a 2e a
n
4.已知等差数列{a n }的前n 项和为S n ,公比为q 的等比数列{b n }的首项是,且
1
2a 1+2q=3,a 2+4b 2=6,S 5=40.
(1)求数列{a n },{b n }的通项公式a n ,b n ;(2)求数列
的前n 项和T n
.
{
1a n a n +1+
1
b n b n +1
}5.已知函数f (x )=,数列{a n }满足:2a n+1-2a n +a n+1a n =0,且a n a n+1≠0.在数列{b n }中,b 1=f (0),且b n =f (a n -
7x +5
x +11).
(1)求证:数列
是等差数列;
{1a n
}(2)求数列{|b n |}的前n 项和T n .
6.已知等比数列{a n }的公比q>1,且a 3+a 4+a 5=28,a 4+2是a 3,a 5的等差中项.数列{b n }满足b 1=1,数列{(b n+1-b n )a n }的前n 项和为2n 2+n.(1)求q 的值;
(2)求数列{b n }的通项公式.
题型练4 大题专项(二)数列的通项、求和问题
1.(1)证明 当n ≥2时,由S n-1-1,S n ,S n+1成等差数列,可知2S n =S n-1-1+S n+1,即S n -S n-1=-1+S n+1-S n ,
即a n =-1+a n+1(n ≥2),则a n+1-a n =1(n ≥2),又a 2-a 1=1,故{a n }是公差为1的等差数列.(2)解 由(1)知等差数列{a n }的公差为1.由S n =0,S n+1=4,得a n+1=4,即a 1+n=4.由S n =0,得na 1+=0,
n (n -1)2即
a 1+=0,解得
n=7.
n -1
22.解 (1)由题意得解得{a 1+3d =7,2(a 1+2d )+a 1+4d =19,{
a 1=1,
d =2.
∴a n =1+2(n-1)=2n-1.
(2)由题意可知b n -a n =2n ,
∴b n =2n +2n-1,
∴T n =(2+22+…+2n )+[1+3+…+(2n-1)],∴T n =2n+1+n 2-2.
3.解 (1)设等差数列{a n }的公差为d ,
∵a 2+a 3=5ln 2.∴2a 1+3d=5ln 2,
又a 1=ln 2,∴d=ln 2.
∴a n =a 1+(n-1)d=n ln 2.
(2)由(1)知a n =n ln 2.
∵=e n ln 2==2n ,
e a
n
e ln 2n
∴{}是以2为首项,2为公比的等比数列.
e a n
∴+…+e
a 1
+e a
2e
a n
=2+22+…+2n =2n+1-2.
∴+…+=2n+1-2.
e a 1+e a 2e a n
4.解 (1)设{a n }公差为d ,由题意得解得
故a n =3n-1,b n =.
{
a 1+2d =8,
a 1+2q =3,a 1+d +2q =6,{a 1=2,d =3,q =12,(12)n
(2)∵+2
2n+1,
1a n a n +1+1b n b n +1=
13(
1
a n -1
a
n +1
)
+
1b n b n +1=13(
1
a n
-1
a
n +1
)∴T n =+…+(22n+3-8)=.
13[(1
2
-15
)+
(15
-
18)(
13n -1
-
13n +2)]+
8(1-4n )1-4
=
13(
12
-1
3n +2)
+1
3
1
3
(
22n +3-
13n +2)
‒
525.(1)证明 ∵2a n+1-2a n +a n+1a n =0,∴,
1
a n +1
‒
1a n =12故数列
是以为公差的等差数列.{1a n
}12
(2)解 ∵b 1=f (0)=5,
∴
=5,7a 1-2=5a 1,
7(a 1-1)+5a 1-1+1
∴a 1
=1,=1+(n-1)·,
1a n 12∴a n =,b n ==7-(n+1)=6-n.
2
n +17a n -2a n 当n ≤6
时,T n =(5+6-n )=;
n
2n (11-n )2当n ≥7
时,T n =15+(1+n-6)=.故
T n =
n -6
2n 2-11n +602{
n (11-n )
2
,n ≤6,n 2-11n +60
2
,n ≥7.6.解 (1)由a 4+2是a 3,a 5的等差中项,得a 3+a 5=2a 4+4,所以a 3+a 4+a 5=3a 4+4=28,解得a 4=8.由a 3+a 5=20,得8=20,
(
q +
1
q
)解得q=2或
q=,
12因为q>1,所以q=2.
(2)设c n =(b n+1-b n )a n ,数列{c n }前n 项和为S n ,
由c n =
解得c n =4n-1.
{
S 1,n =1,S n -S n -1,n ≥2,
由(1)可知a n =2n-1,所以b n+1-b n
=(4n-1)·.(12)
n -1
故b n -b n-1
=(4n-5)·,n ≥2,
(12)
n -2
b n -b 1=(b n -b n-1)+(b n-1-b n-2)+…+(b 3-b 2)+(b 2-b 1)=(4n-5)·+(4n-9)·+…+7·+3.
(12)
n -2(12)
n -31
2设
T n =3+7·+11·+…+(4n-5)·,n ≥2,
12(12)2(12)
n -2
T n =3·+7·+…+(4n-9)·+(4n-5)·,1212(12)2(12)n -2(12)
n -1所以T n =3+4·+4·+…+4·-(4n-5)·,
1212(12)2(12)n -2(12)
n -1因此T n =14-(4n+3)·
,n ≥2,
(12)
n -2
又b 1=1,所以b n =15-(4n+3)·
.
(12)
n -2。
