高考数学一轮总复习考点规范练28 数列求和 (2).docVIP

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考点规范练28数列求和

1.在数列{an}中,如果a1=1,a2=2,an+2-an=1+(-1)n,那么S100的值为()

A.2500 B.2600 C.2700 D.2800

答案:B

解析:当n为奇数时,an+2-an=0,所以an=1;

当n为偶数时,an+2-an=2,所以an=n,

故an=1

于是S100=50+(2+100

2.已知数列{an}满足a1=1,且对任意的n∈N*都有an+1=a1+an+n,则数列1a

A.100101 B.99100 C.101

答案:D

解析:∵an+1=a1+an+n,a1=1,

∴an+1-an=1+n.

∴an-an-1=n(n≥2).

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

∴1an=

∴数列1an的前100项和为2×(1-12+12

3.已知函数f(x)=xa的图象过点(4,2),令an=1f(n+1)+f(n),n∈N

A.1020-1 B.1020+1

C.1021-1 D.1021+1

答案:C

解析:由f(4)=2,可得4a=2,

解得a=12

则f(x)=x1

即an=1f

则S1020=a1+a2+a3+…+a1020=(2-1)+(3-2)+(4-

4.已知数列{an}满足an+1+(-1)nan=2n-1,则数列{an}的前60项和为()

A.3690 B.3660

C.1845 D.1830

答案:D

解析:∵an+1+(-1)nan=2n-1,

∴当n=2k(k∈N*)时,a2k+1+a2k=4k-1;①

当n=2k-1(k∈N*)时,a2k-a2k-1=4k-3;②

当n=2k+1(k∈N*)时,a2k+2-a2k+1=4k+1.③

由①-②得a2k+1+a2k-1=2,

∴(a1+a3)+(a5+a7)+…+(a57+a59)=2×15=30.

由①+③得a2k+a2k+2=8k,

∴(a2+a4)+(a6+a8)+…+(a58+a60)=15×8+15×142

∴a1+a2+…+a60=30+1800=1830.

5.已知等差数列{an}中,a5=π2.若函数f(x)=sin2x+1,设yn=f(an),则数列{yn}的前9项和为

答案:9

解析:由题意,得yn=sin(2an)+1,

故数列{yn}的前9项和为sin2a1+sin2a2+sin2a3+…+sin2a8+sin2a9+9.

由a5=π2,得sin2a5

∵a1+a9=2a5=π,

∴2a1+2a9=4a5=2π,

∴2a1=2π-2a9,

∴sin2a1=sin2π-2a

由倒序相加可得12(sin2a1+sin2a2+sin2a3+…+sin2a8+sin2a9+sin2a1+sin2a2+sin2a3+…+sin2a8+sin2a9

∴y1+y2+y3+…+y8+y9=9.

6.已知数列{an}的前n项和为Sn,且2Sn+3=3an.设bn=(n+1)log?3an,并记Tn=1b1+1b2+1b3

答案:3n4021

解析:当n=1时,可得2S1+3=2a1+3=3a1,

解得a1=3;

当n≥2时,2Sn-1+3=3an-1,

则2Sn+3-(2Sn-1+3)=3an-3an-1,即2an=3an-3an-1,

即an=3an-1,则数列{an}是首项为3,公比为3的等比数列,

故数列{an}的通项公式为an=3n.

则bn=(n+1)log?33n=(n+1)log?3(

即1b

可得Tn=12[1-12+

故T4020=12

7.(新高考Ⅱ,18)已知{an}为等差数列,bn=an-6,n为奇数,2an,n为偶数.记S

(1)求{an}的通项公式;

(2)证明:当n5时,TnSn.

(1)解:设等差数列{an}的公差为d.

由bn=an-6,n为奇数,2an,n为偶数,得b1=a1-6,b

则由S4=32,T3=16,得4

解得a

所以an=a1+(n-1)d=2n+3.

(2)证明:由(1)可得Sn=n[5+(

当n为奇数时,Tn=a1-6+2a2+a3-6+2a4+a5-6+2a6+…+an-2-6+2an-1+an-6=(-1+14)+(3+22)+(7+30)+…+[(2n-7)+(4n+2)]+2n-3=[-1+3+…+(2n-7)+(2n-3)]+[14+22+…+(4n+2)]=n+12(-

当n5时,Tn-Sn=3n2+5n-102-(n2+4n)=

当n为偶数时,Tn=a1-6+2a2+a3-6+2a4+a5-6+2a6+…+an-1-6+2an=(-1+14)