高考解答题专项训练3

高考解答题专项训练(三) 数列

1.(2019·咸阳模拟)在△ABC中,角A,B,C的对边分别为a,b,c,B=60°,三边a,b,c成等比数列,且面积为43,在等差数列{an}中,a1=4,公差为b.

(1)求数列{an}的通项公式;

16(2)数列{cn}满足cn=,设Tn为数列{cn}的前n项和,求Tn.

anan+1解:(1)由a,b,c成等比数列得b2=ac, 1

因为S△ABC=43=2acsinB,所以b=4,

所以{an}是以4为首项,以4为公差的等差数列, 其通项公式为an=4n.

111

(2)由(1)可得cn==-,

n?n+1?nn+1

?11?1??11??1n-??????Tn=1-2+2-3+…+nn+1=1-=. ????n+1n+1??

2.(2019·安徽淮南一模)已知数列{an}为等差数列,且a3=5,a5=9,数列{bn}21

的前n项和为Sn=3bn+3.

(1)求数列{an}和{bn}的通项公式; (2)设cn=an|bn|,求数列{cn}的前n项和Tn. 解:(1)∵数列{an}为等差数列,且a3=5,a5=9, a5-a39-5∴d==2=2,

5-3∴a1=a3-2d=5-4=1, ∴an=1+(n-1)×2=2n-1.

21

∵数列{bn}的前n项和为Sn=3bn+3, 21

∴n=1时,S1=3b1+3, 由S1=b1,解得b1=1,

22

当n≥2时,bn=Sn-Sn-1=3bn-3bn-1,

∴bn=-2bn-1,∴{bn}是首项为1,公比为-2的等比数列,∴bn=(-2)n-1. (2)cn=an|bn|=(2n-1)·2n-1,

∴数列{cn}的前n项和Tn=1×1+3×2+5×22+…+(2n-1)×2n-1, ∴2Tn=1×2+3×22+5×23+…+(2n-1)×2n, 两式相减,得:

n

2-2

-Tn=1+2(2+22+…+2n-1)-(2n-1)·2n=1+2×-(2n-1)·2n=1+2n+

1-21

-4-(2n-1)·2n=-3+(3-2n)·2n,

∴Tn=(2n-3)·2n+3.

3.已知等差数列{an}的公差为2,前n项和为Sn,且S1,S2,S4成等比数列. (1)求数列{an}的通项公式; (2)令bn=(-1)n-1

4n

,求数列{bn}的前n项和Tn. anan+1

2×1

解:(1)因为S1=a1,S2=2a1+2×2=2a1+2, 4×3

S4=4a1+2×2=4a1+12, 所以由题意得(2a1+2)2=a1(4a1+12), 解得a1=1, 所以an=2n-1. (2)bn=(-1)=(-1)=(-1)

n-1n-1

4n

anan+1

4n

?2n-1??2n+1?

?11?

+??. 2n-12n+1??

n-1

当n为偶数时,

?11??11?1??11??1

++Tn=?1+3?-?3+5?+…+?2n-32n-1?-?2n-12n+1?=1-=

????2n+1????

2n. 2n+1

当n为奇数时,

?11??11?1??11??1

++Tn=?1+3?-?3+5?+…-?2n-32n-1?+?2n-12n+1?=1+=

????2n+1????

2n+2. 2n+1

?所以T=?2n

?2n+1,n为偶数.

n*

2n+2

,n为奇数,2n+1

?2n+1+?-1?n-1??或Tn=?

2n+1??

4.(2019·烟台模拟)已知二次函数f(x)=ax2+bx的图象过点(-4n,0),且f′(0)

?1?

=2n,n∈N,数列{an}满足=f′?a?,且a1=4.

?n?an+1

1

(1)求数列{an}的通项公式;

(2)记bn=anan+1,求数列{bn}的前n项和Tn. 解:(1)f′(x)=2ax+b, 由题意知b=2n,16n2a-4nb=0, 112

∴a=2,则f(x)=2x+2nx,n∈N*.

?1?

数列{an}满足=f′?a?,

?n?an+1

1

又f′(x)=x+2n, ∴

111

=a+2n,∴-a=2n, an+1nan+1n1

11

由叠加法可得a-4=2+4+6+…+2(n-1)=n2-n,

n4

化简可得an=(n≥2),

?2n-1?2当n=1时,a1=4也符合, 4*

∴an=2(n∈N). ?2n-1?

?11?4

-?, (2)∵bn=anan+1==2?

?2n-1??2n+1??2n-12n+1?

∴Tn=b1+b2+…+bn =a1a2+a2a3+…+anan+1

???11??1??11?

=2??1-3?+?3-5?+…+?2n-1-2n+1??

?????????1?4n1-??=22n+1?=2n+1. ?

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