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课时跟踪检测(十二) 数列求和 层级一 n 学业水平达标 ) 1.已知 an=(-1) ,数列{an}的前 n 项和为 Sn,则 S9 与 S10 的值分别是( A.1,1 C.1,0 解析:选 D B.-1,-1 D.-1,0 S9=-1+1-1+1-1+1-1+1-1=-1, S10=S9+a10=-1+1=0. 2.数列{an}的通项公式是 an= A.11 C.120 解析:选 C ∵an= 1 1 n+ n+1 ,若前 n 项和为 10,则项数为( B.99 D.121 ) n+ n+1 = n+1- n, ∴Sn=a1+a2+…+an =( 2-1)+( 3- 2)+…+( n+1- n) = n+1-1, 令 n+1-1=10,得 n=120. 3.已知数列{an},a1=2,an+1-2an=0,bn=log2an,则数列{bn}的前 10 项和等于( A.130 C.55 解析:选 C B.120 D.50 在数列{an}中,a1=2,an+1-2an=0,即 ) an+1 =2,所以数列{an}是以 2 为首 an 项,2 为公比的等比数列. 所以 an=2×2 n-1 =2 . n 所以 bn=log22 =n. 则数列{bn}的前 10 项和为 1+2+…+10=55.故选 C. 4.在数列{an}中,已知 Sn=1-5+9-13+17-21+…+(-1) n-1 n (4n-3),则 S15+S22- S31 的值( A.13 C.46 ) B.-76 D.76 ∵S15=(-4)×7+(-1) (4×15-3)=29. 14 解析:选 B S22=(-4)×11=-44. S31=(-4)×15+(-1)30(4×31-3)=61. ∴S15+S22-S31=29-44-61=-76. 5.数列 1,1+2,1+2+2 ,…,1+2+2 +…+2 A.2 -101 C.2 -99 解析:选 A 2 100 100 2 2 n-1 ,…的前 99 项和为( ) B.2 -101 D.2 -99 由数列可知 an=1+2+2 +…+2 n-1 99 99 = 1-2 n =2 -1,所以,前 99 项的和为 1-2 -2 1-2 99 n S99=(2-1)+(22-1)+…+(299-1)=2+22+…+299-99= -99=2 -101. ? 100 6.已知等比数列{an}的公比 q≠1,且 a1=1,3a3=2a2+a4,则数列? ________. ?anan+1? 1 ? ?的前 4 项和为 解析:∵等比数列{an}中,a1=1,3a3=2a2+a4,∴3q =2q+q .又∵q≠1,∴q=2,∴an =2 n-1 2 3 ,∴ ? 1 ? 1 1 ?1?2n-1 ?是首项为 ,公比为 的等比数列, =? ? ,即? anan+1 ?2? 2 4 ?anan+1? 1 ? 1 ? 2? ?的前 4 项和为 ∴数列? ?anan+1? 1? ?1?4? ?1-?4? ? ? ? ? 85 = . 1 128 1- 4 85 答案: 128 7.等比数列{an}的前 n 项和为 Sn,若 =3,则 =________. 解析: =3,故 q≠1, ∴ S6 S3 S9 S6 S6 S3 a1 3 -q 1-q 6 × a1 1-q 3 -q =1+q =3, 3 即 q =2. 所以 = 答案: 7 3 3 S9 a1 -q9 1-q 1-2 7 × = 6 2= . S6 1-q a1 1-q 1-2 3 8.对于数列{an},定义数列{an+1-an}为数列{an}的“差数列”,若 a1=2,{an}的“差数 列”的通项公式为 2 ,则数列{an}的前 n 项和 Sn=________. 解析:∵an+1-an=2 , ∴an=(an-an-1)+(an-1-an-2)+…+(a2-a1)+a1 =2 n-1 n n +2 n-2 +…+2 +2+2= 2 2-2 n n +2=2 -2+2=2 . 1-2 n ∴Sn= 2-2 n+1 =2 -2. 1-2 n+1 n+1 答案:2 -2 2 9.已知{an}是递增的等差数列,a1=2,a2=a4+8. (1)求数列{an}的通项公式; (2)若 bn=an+2an,求数列{bn}的前 n 项和 Sn. 解:(1)设数列{an}的公差为 d,d>0.由题意得(2+d) =2+3d+8,解得 d=2. 故 an=a1+(n-1)·d=2+(n-1)·2=2n. (2)∵bn=an+2an=2n+2 , ∴Sn=b1+b2+…+bn =(2+2 )+(4+2 )+…+(2n+2 ) =(2+4+…+2n)+(2 +2 +…+2 ) = +2n 2 4 2 4 2n 2 4 2n 2n 2 n + -4 1-4 n n+1 =n(n+1)+ -4 . 3 10.在等差数列{an}中,a3=4,a7=8. (1)求数列{an}的通项公式 an; (2)令 bn= an 2 n-1 ,求数列{bn}的前 n 项和 Tn. 解:(1)因为 d= (2)bn= a7-a3 7-3 , =1,所以 an=a3+(n-3)d=n+1. an 2 n-1 = n+1 2 n-1 Tn=b1+b2+…+bn=2+ + 2+…+ 1 2 3 n n+1 Tn= + 2+…+ n-1+ n , 2 2 2 2 2 3 2 4 2 n+1 2 n-1 . ① ② 1 1 1 1 n+1 由①-②得 Tn=2+ + 2+…+ n-1- n 2 2 2 2 2 1 ? n+1 ? 1 1 =?1+ + 2+…+ n-1?+1- n 2 ? 2 ? 2 2 1 n

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