【精品】高中数学一轮复习基础讲解数列求和(含解析)

数列”的通项公式为2n,则数列{an}的前n项和Sn=________.

解析:∵an+1-an=2n,

∴an=(an-an-1)+(an-1-an-2)+…+(a2-a1)+a1 =2

n-1

+2

n-2

2-2n

+…+2+2+2=+2=2n-2+2=2n.

1-2

2

2-2n1n+1

∴Sn==2-2.

1-2答案:2n1-2

?1?

9.已知等比数列{an}中,a1=3,a4=81,若数列{bn}满足bn=log3an,则数列?bb?的

?nn+1?

前n项和Sn=________.

a4--

解析:设等比数列{an}的公比为q,则=q3=27,解得q=3.所以an=a1qn1=3×3n1

a1

=3n,故bn=log3an=n,

1111所以==-.

bnbn+1n?n+1?nn+1

?1?111111n则数列?bb?的前n项和为1-+-+…+-=1-=. 223nn+1n+1n+1?nn+1?

答案:

n

n+1

10.(2013·唐山统考)在等比数列{an}中,a2a3=32,a5=32. (1)求数列{an}的通项公式;

(2)设数列{an}的前n项和为Sn,求S1+2S2+…+nSn. 解:(1)设等比数列{an}的公比为q,依题意得

?a1q2=32,?a1q·?解得a1=2,q=2, 4=32,?aq?1

故an=2·2n1=2n.

(2)∵Sn表示数列{an}的前n项和, 2?1-2n?∴Sn==2(2n-1),

1-2

∴S1+2S2+…+nSn=2[(2+2·22+…+n·2n)-(1+2+…+n)]=2(2+2·22+…+n·2n)-n(n+1),

设Tn=2+2·22+…+n·2n,① 则2Tn=22+2·23+…+n·2n1,② ①-②,得 -Tn

=2+22+…+2n-n·2n1=

2?1-2n?++

-n·2n1=(1-n)2n1-2, 1-2

∴Tn=(n-1)2n1+2,

∴S1+2S2+…+nSn=2[(n-1)2n1+2]-n(n+1) =(n-1)2n2+4-n(n+1).

11.(2012·长春调研)已知等差数列{an}满足:a5=9,a2+a6=14. (1)求{an}的通项公式;

(2)若bn=an+qan(q>0),求数列{bn}的前n项和Sn.

??a1+4d=9,解:(1)设数列{an}的首项为a1,公差为d,则由a5=9,a2+a6=14,得?

?2a1+6d=14,???a1=1,

解得?所以{an}的通项an=2n-1.

?d=2,?

(2)由an=2n-1得bn=2n-1+q2n1.

当q>0且q≠1时,Sn=[1+3+5+…+(2n-1)]+(q1+q3+q5+…+q2n1)=n2+q?1-q2n?;

1-q2

当q=1时,bn=2n,则Sn=n(n+1). 所以数列{bn}的前n项和

n?n+1?,q=1,??

Sn=?2q?1-q2n?

n+,q>0,q≠1.?1-q2?

212.(2012·“江南十校”联考)若数列{an}满足:a1=,a2=2,3(an+1-2an+an-1)=2.

3(1)证明:数列{an+1-an}是等差数列;

11115

(2)求使+++…+>成立的最小的正整数n.

a1a2a3an2解:(1)由3(an+1-2an+an-1)=2可得:

22

an+1-2an+an-1=,即(an+1-an)-(an-an-1)=,

33

42

故数列{an+1-an}是以a2-a1=为首项,为公差的等差数列.

33422

(2)由(1)知an+1-an=+(n-1)=(n+1),

333

21

于是累加求和得an=a1+(2+3+…+n)=n(n+1),

33111

∴=3?n-n+1?, an??

111135∴+++…+=3->,∴n>5, a1a2a3ann+12∴最小的正整数n为6.

1.已知数列{an}的前n项和Sn=n2-6n,则{|an|}的前n项和Tn=( ) A.6n-n2

2??6n-n?1≤n≤3?

C.?2 ?n-6n+18?n>3??

B.n2-6n+18

2??6n-n ?1≤n≤3?

D.?2 ?n-6n ?n>3??

解析:选C ∵由Sn=n2-6n得{an}是等差数列,且首项为-5,公差为2. ∴an=-5+(n-1)×2=2n-7, ∴n≤3时,an<0,n>3时,an>0,

2??6n-n?1≤n≤3?,

∴Tn=?2

?n-6n+18?n>3?.?

2.(2012·成都二模)若数列{an}满足a1=2且an+an-1=2n+2n1,Sn为数列{an}的前n项和,则log2(S2 012+2)=________.

解析:因为a1+a2=22+2,a3+a4=24+23,a5+a6=26+25,….所以S2 012=a1+a2+a3+a4+…+a2 011+a2 012

=21+22+23+24+…+22 011+22 012 2?1-22 012?2 013

==2-2.

1-2

故log2(S2 012+2)=log222 013=2 013. 答案:2 013

3.已知递增的等比数列{an}满足:a2+a3+a4=28,且a3+2是a2,a4的等差中项. (1)求数列{an}的通项公式;

1

(2)若bn=anlogan,Sn=b1+b2+…+bn,求Sn.

2解:(1)设等比数列{an}的首项为a1,公比为q. 依题意,有2(a3+2)=a2+a4, 代入a2+a3+a4=28,得a3=8. ∴a2+a4=20.

?3???q=2,?a1q+a1q=20,?q=2,

∴?解得?或??a3=a1q2=8,???a1=2,?

1

?a1=32.

又{an}为递增数列,

??q=2,∴?∴an=2n. ??a1=2.

1(2)∵bn=2n·log2n=-n·2n,

2

∴-Sn=1×2+2×22+3×23+…+n×2n.①

∴-2Sn=1×22+2×23+3×24+…+(n-1)×2n+n×2n1.② ①-②得Sn=2+22+23+…+2n-n·2n1 2?1-2n?+=-n·2n1

1-2=2n1-n·2n1-2. ∴Sn=2n1-n·2n1-2.

1.已知{an}是公差不为零的等差数列,a1=1,且a1,a3,a9成等比数列. (1)求数列{an}的通项; (2)求数列{2an}的前n项和Sn.

1+2d1+8d解:(1)由题设知公差d≠0,由a1=1,a1,a3,a9成等比数列得=,

11+2d解得d=1或d=0(舍去), 故{an}的通项an=1+(n-1)×1=n. (2)由(1)知2an=2n, 由等比数列前n项和公式得 Sn

=2+22+23+…+2n=

2?1-2n?n+1

=2-2. 1-2

3

2.设函数f(x)=x3,在等差数列{an}中,a3=7,a1+a2+a3=12,记Sn=f(an+1),令

?1?

bn=anSn,数列?b?的前n项和为Tn.

?n?

(1)求{an}的通项公式和Sn; 1

(2)求证:Tn<.

3

解:(1)设数列{an}的公差为d,由a3=a1+2d=7,a1+a2+a3=3a1+3d=12,解得a1

=1,d=3,则an=3n-2.

3

∵f(x)=x3,∴Sn=f(an+1)=an+1=3n+1. (2)证明:∵bn=anSn=(3n-2)(3n+1), 11111

∴==?3n-2-3n+1?. bn?3n-2??3n+1?3??111∴Tn=++…+ b1b2bn

111111

=?1-4+4-7+…+3n-2-3n+1? 3??

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