(新课标)2017高考数学大一轮复习第五章数列35数列求和课时作业理

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即an=3

n-1

??3,n=1,

,所以an=?n-1

?3,n>1.?

1

(2)因为anbn=log3an,所以b1=,

3当n>1时,bn=3

1-nlog33

n-1

=(n-1)·3

1-n,

1

所以T1=b1=;

3

1-1-21-n当n>1时,Tn=b1+b2+b3+…+bn=+[1×3+2×3+…+(n-1)×3],

3所以,3Tn=1+[1×3+2×3+…+(n-1)×3两式相减,得

20-1-22-n1-n2Tn=+(3+3+3+…+3)-(n-1)×3

321-3136n+31-n=+=--1-(n-1)×3n, 31-362×3136n+3所以Tn=-n.

124×3经检验,n=1时也适合. 136n+3

综上可得Tn=-n.

124×3

1-n0

-1

2-n],

1111+2+2+…+的值为( ) 2

2-13-14-1(n+1)-1

2

1.A.

n+1

2(n+2)

3n+1B.- 42(n+2)1?31?1+C.-??

42?n+1n+2?1?31?1-D.-??

42?n+1n+2?

111

解析:∵=2= 2

(n+1)-1n+2nn(n+2)1?1?1

=?-?, 2?nn+2?∴

1111+2+2+…+ 2

2-13-14-1(n+1)-1

2

11?1?11111

=?1-+-+-+…+-

32435nn+2?2??11?1?3

-=?-?

2?2n+1n+2?1?31?1+=-??.

42?n+1n+2?答案:C 2.定义

np1+p2+…+pn为n个正数p1,p2,…,pn的“均倒数”.若已知正项数列{an}

的前n项的“均倒数”为

A.C.1 1110 11

1an+1111,又bn=,则++…+=( ) 2n+14b1b2b2b3b10b11

B.1

12

11D. 12

12

=,即Sn=n(2n+1)=2n+n,当n=1时,a1=S1

a1+a2+…+an2n+1

解析:依题意有

n=3;当n≥2时,an=Sn-Sn-1=4n-1,a1=3满足该式.则an=4n-1,bn=1=

an+1

4

=n.因为

bnbn+1

1111111111110

=-,所以++…+=1-+-+…+-=.

n(n+1)nn+1b1b2b2b3b10b11223101111

答案:C

??1?11??11??是等差数列,3.设数列{an}是首项为1的等比数列,若?则?+?+?+??2a1a2??2a2a3??2an+an+1?

+…+?

?1+1?的值等于( )

??2a2 012a2 013?

B.2 013 D.3 019

A.2 012 C.3 018

??1112

?是等差数列,解析:设数列{an}的公比为q,若?则由+=,

2a1+a22a3+a42a2+a3?2an+an+1?

112113?11??11?+2+=,故?+?+?+?3=2,解得q=1,则an=1,从而2+q2q+q2q+q2anan+12?2a1a2??2a2a3?

+…+?

?1+1?=2 012×3=3 018.

?2?2a2 012a2 013?

*

答案:C

4.(2016·韶关模拟)已知数列{an}的首项a1=1,前n项和为Sn,an+1=2Sn+1,n∈N. (1)求数列{an}的通项公式;

?bn?9

(2)设bn=log3an+1,求数列??的前n项和Tn,并证明:1≤Tn<.

4?an?

解:(1)由an+1=2Sn+1, 得an=2Sn-1+1(n≥2),

两式相减得an+1-an=2(Sn-Sn-1)=2an, 故an+1=3an(n≥2),

所以当n≥2时,{an}是以3 为公比的等比数列. 因为a2=2S1+1=2a1+1=3,=3,

所以{an}是首项为1,公比为3的等比数列,an=3(2)证明:由(1)知an=3

n-1

n-1

a2

a1

.

,故bn=log3an+1=log33=n,

n1?n-1bnn?==n·??,

an3n-1?3?

1?21?31?n-11???Tn=1+2×+3×??+4×??+…+n×??,①

3?3??3??3?

1?21?31?n-11?n11????Tn=1×+2×??+3×??+…+(n-1)×??+n×??.②

33?3??3??3??3?

?1?1-?3?1?n-11?n1?n21?1?2?1?3?????①-②,得Tn=1++??+??+…+??-n×??=-n×??,

33?3?1?3??3??3??3?

1-3

9?93??1?n所以Tn=-?+n???.

4?42??3?

n?93??1?因为?+n???>0, ?42??3?

9?93??1?n9

所以Tn=-?+n???<.

4?42??3?4又因为Tn+1-Tn=

nn+1

3

n>0,

所以数列{Tn}单调递增, 所以(Tn)min=T1=1, 9

所以1≤Tn<. 4


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