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专题突破练 12 求数列的通项及前 n 项和
1.(2021·湖南长郡中学月考)已知等比数列{a}的各项均为正数,且2a+3a=1,a2=9aa.
n 1 2 3 2 6
(1)求数列{a}的通项公式;
n
{1 }
(2)设b=log a+log a+…+log a,求数列 的前n项和T.
n 3 1 3 2 3 n b n
n
2.(2021·山东威海期末)已知等差数列{a}的前n项和为S,且满足a=8,S=2a.
n n 3 5 7
(1)求数列{a}的通项公式;
n
(2)若数列{b}满足b=a cos nπ+2n+1,求数列{b}的前2n项和T .
n n n n 2n3.(2021·东北三省四市联考)已知等差数列{a}的前n项和为S,S=25,且a-1,a+1,a+3成等比数列.
n n 5 3 4 7
(1)求数列{a}的通项公式;
n
(2)若b=(-1)na+1,T 是数列{b}的前n项和,求T .
n n n n 2n
4.(2021·陕西西安铁一中月考)已知数列{a}是公差不为0的等差数列,且a=3,a,a,a 成等比数列.
n 2 1 2 5
(1)求数列{a}的通项公式;
n
1
(2)设S 为数列{a+2}的前n项和,b= ,求数列{b}的前n项和T.
n n n n n
S
n5.(2021·广东揭阳检测)已知等差数列{a}与正项等比数列{b}满足a=b =3,且b-a,20,a+b 既是等
n n 1 1 3 3 5 2
差数列,又是等比数列.
(1)求数列{a}和{b}的通项公式;
n n
(2)在以下三个条件中任选一个,补充在下面问题中,并完成求解.
1 2(a +3)
①c= +(-1)nb,②c=a ·b,③c= n .
n a ·a n n n n n a a b
n n+1 n n+1 n+1
若 ,求数列{c}的前n项和S.
n n
6.(2021·山东菏泽一模)已知等比数列{a}的前n项和为S,且a =2S+2,数列{b}满足b=2,
n n n+1 n n 1
(n+2)b=nb .
n n+1
(1)求数列{a}和{b}的通项公式;
n n
(2)在a 与a 之间插入n个数,使这n+2个数组成一个公差为c 的等差数列,求数列{bc}的前n项和
n n+1 n n n
T.
n7.(2021·广东广州检测)已知数列{a}满足a=1,a =3a+3n+1.
n 1 n+1 n
{a }
(1)求证:数列 n 是等差数列;
3n
(2)求数列{a}的通项公式;
n
S 3n 7
(3)设数列{a}的前n项和为S,求证: n> − .
n n
3n 2 4专题突破练 12 求数列的通项及前 n 项和
1 1
1.解: (1)设等比数列{a
n
}的公比为q(q>0),由 a2=9a
2
a
6
,得 a2=9 a2,所以q2= ,所以q= .
3 3 4 9 3
1 1
由2a +3a =1,得2a +3a · =1,所以a = .
1 2 1 1 1
3 3
1
故数列{a }的通项公式为a = .
n n
3n
n(n+1) 1 2
(2)因为b =log a +log a +…+log a =-(1+2+…+n)=- ,所以 =- =-2
n 3 1 3 2 3 n
2 b n(n+1)
n
(1
-
1 ).
n n+1
所以T = 1 + 1 +…+ 1 =-2 ( 1- 1) + (1 - 1) +…+ (1 - 1 ) =- 2n .
n b b b 2 2 3 n n+1 n+1
1 2 n
{1 } 2n
所以数列 的前n项和T =- .
b n n+1
n
2.解: (1)设{a }的公差为d,依题意,¿
n
{a =2,
解得 1 所以a =2+3(n-1)=3n-1.
d=3. n
(2)因为b =a cos nπ+2n+1=(-1)na +2n+1=(-1)n·(3n-1)+2n+1,
n n n
22(1-22n)
所以T =(a -a )+(a -a )+…+(a -a )+(22+23+…+22n+1)=3n+ =3n+22n+2-4.
2n 2 1 4 3 2n 2n-1
1-2
5(a +a )
3.解: (1)由题意可知S = 1 5 =5a =25,所以a =5.
5 3 3
2
设等差数列{a }的公差为d,由a -1,a +1,a +3成等比数列,
n 3 4 7
可得(6+d)2=4(8+4d),
整理得d2-4d+4=0,解得d=2.
所以a =a +(n-3)d=2n-1.
n 3
(2)因为b =(-1)na +1=(-1)n(2n-1)+1,
n n
所以T =(-1+1)+(3+1)+(-5+1)+(7+1)+…+[-(4n-3)+1]+(4n-1+1)=4n.
2n
4.解: (1)设等差数列{a }的公差为d(d≠0),
n
则由题意,可知{ a +d=3, 解得{a =1,
1 1
(a +d)2=a (a +4d), d=2.
1 1 1
∴a =1+2(n-1)=2n-1.
n
(2)由(1)得a +2=2n+1,∴S =(a +2)+(a +2)+(a +2)+…+(a +2)+(a +2)=3+5+7+…
n n 1 2 3 n-1 n
(2n+1+3)n
+(2n-1)+(2n+1)= =n2+2n.
2∴b = 1 = 1 = 1 = 1(1 - 1 ) .
n
S n2+2n n(n+2) 2 n n+2
n
∴T =b +b +b +…+b +b = 1( 1- 1) + (1 - 1) + (1 - 1) +…+ ( 1 - 1 ) + 1 − 1
n 1 2 3 n-1 n 2 3 2 4 3 5 n-1 n+1 n n+2
= 1( 1+ 1 - 1 - 1 ) = 3 − 2n+3 .
2 2 n+1 n+2 4 2(n+1)(n+2)
5.解: (1)设等差数列{a }的公差为d,等比数列{b }的公比为q(q>0),由已知得20=b -
n n 3
a =a +b ,即20=3q2-(3+2d),20=(3+4d)+3q,解得d=2,q=3,所以a =2n+1,b =3n.
3 5 2 n n
(2)若选择①,
则c = 1 +(-1)nb = 1 +(-3)n= 1( 1 - 1 ) +(-3)n,
n a ·a n (2n+1)(2n+3) 2 2n+1 2n+3
n n+1
所以S =c +c +…+c = 1 × (1 - 1) +(-3)1+ 1 × (1 - 1) +(-3)2+…+ 1( 1 - 1 ) +
n 1 2 n
2 3 5 2 5 7 2 2n+1 2n+3
(-3)n= 1(1 - 1 ) + -3[1-(-3)n] = n − 3[1-(-3)n] .
2 3 2n+3 1+3 3(2n+3) 4
若选择②,
则c =a ·b =(2n+1)3n,
n n n
所以S =c +c +…+c =3×3+5×32+…+(2n+1)3n,
n 1 2 n
3S =3×32+5×33+…+(2n+1)3n+1,
n
两式相减得-2S =32+2×32+2×33+…+2×3n-(2n+1)3n+1=-2n·3n+1,所以S =n·3n+1.
n n
若选择③,
2(a +3) 2(2n+4) 1 1
则c = n = = − ,
n a a b (2n+1)(2n+3)3n+1 (2n+1)3n (2n+3)3n+1
n n+1 n+1
( 1 1 ) 1 1
所以S =c +c +…+c = - + − +…+
n 1 2 n 3×3 5×32 5×32 7×33
[ 1 1 ] 1 1
- = − .
(2n+1)3n (2n+3)3n+1 9 (2n+3)3n+1
6.解: (1)设等比数列{a }的公比为q,由a =2S +2,可得a =2S +2(n≥2),
n n+1 n n n-1
两式相减得a -a =2S -2S =2a ,
n+1 n n n-1 n
整理得a =3a ,可知q=3.
n+1 n
令n=1,则a =2a +2,即3a =2a +2,解得a =2.
2 1 1 1 1
故a =2·3n-1.
n
b n+2
由b =2,(n+2)b =nb ,得 n+1= ,
1 n n+1
b n
n
b b b n+1 n 3
则当n≥2时,b = n · n-1·…· 2·b = · ·…· ×2=n(n+1).
n 1
b b b n-1 n-2 1
n-1 n-2 1
又b =2满足上式,所以b =n(n+1).
1 n
(2)若在a 与a 之间插入n个数,使这n+2个数组成一个公差为c 的等差数列,
n n+1 n4·3n-1
则a -a =(n+1)c ,即2·3n-2·3n-1=(n+1)c ,整理得c = ,所以b c =4n·3n-1,
n+1 n n n n n n
n+1
所以T =b c +b c +b c +…+b c +b c =4×1×30+4×2×31+4×3×32+…
n 1 1 2 2 3 3 n-1 n-1 n n
+4·(n-1)3n-2+4·n·3n-1=4[1×30+2×31+3×32+…+(n-1)3n-2+n·3n-1],
3T =4[1×31+2×32+…+(n-1)3n-1+n·3n],
n
(1-3n
)
两式相减得-2T =4(30+31+32+…+3n-1-n·3n)=4 -n·3n =(2-4n)·3n-2,所以
n
1-3
T =(2n-1)3n+1.
n
a a a a
7.(1)证明: 由a =3a +3n+1,得 n+1= n+1,即 n+1− n=1.
n+1 n
3n+1 3n 3n+1 3n
a 1 {a } 1
又 1= ,所以数列 n 是以 为首项,1为公差的等差数列.
3 3 3n 3
a 1 2
(2)解: 由(1)得 n= +(n-1)×1=n- ,
3n 3 3
所以a = ( n- 2) ·3n.
n
3
(3)证明: 由(2)得S = ( 1- 2) ×31+ ( 2- 2) ×32+…+ [ (n-1)- 2] ×3n-1+ ( n- 2) ×3n,
n
3 3 3 3
( 2) ( 2) 2 ( 2)
3S = 1- ×32+ 2- ×33+…+ (n-1)- ×3n+ n- ×3n+1,
n
3 3 3 3
两式相减得2S = ( n- 2) ×3n+1- 3n+1-9 -1= n- 7 ×3n+1+ 7 ,
n
3 2 6 2
故S n =(n 2 - 1 7 2 )3n+1+ 7 4 ,从而S n= (n 2 - 1 7 2 ) 3n+1 + 7 = (3n - 7) + 7 > 3n − 7.
3n 3n 4×3n 2 4 4×3n 2 4
