发布时间 : 星期五 文章2.5第二课时知能演练轻松闯关更新完毕开始阅读6b7dfa2385254b35eefdc8d376eeaeaad0f31634
1.已知数列{an}的前n项和Sn=1-5+9-13+17-21+…+(-1)n1(4n-3),则S15
=( )
A.-29 B.29 C.30 D.-30
解析:选B.S15=1-5+9-13+…+57=-4×7+57=29.
1
2.数列{an}的通项公式是an=,若前n项和为10,则项数为( )
n+n+1
A.11 B.99 C.120 D.121
1
解析:选C.∵an==n+1-n,
n+n+1∴Sn=a1+a2+…+an=2-1+3-2+…+=
n+1-1=10, ∴n+1=121,
∴n=120. 3.(2019·滨州调研)数列9,99,999,9 999,…,的前n项和等于( )
10?10n-1?n
A.10-1 B.-n
9
1010C.(10n-1) D.(10n-1)+n 99解析:选B.an=10n-1, ∴Sn=a1+a2+…+an
=(10-1)+(102-1)+…+(10n-1)
10?10n-1?=(10+102+…+10)-n=-n.
9
111
4.数列1,,,…,的前n项和为( )
1+21+2+31+2+…+n
2n2nA. B. 2n+1n+1n+2nC. D. n+12n+1
211
解析:选B.该数列的通项为an=,分裂为两项差的形式为an=2(-),令n
nn+1n?n+1?
n
-
n+1-n
=1,2,3,…,
1111111
则Sn=2(1-+-+-+…+-),
22334nn+1
12n
∴Sn=2(1-)=.
n+1n+1
11212312341
5.已知数列{an}={,+,++,+++,…},那么数列{bn}={}前
2334445555anan+1
n项的和为( )
111
A.4(1-) B.4(-) 2n+1n+1
111
C.1- D.- 2n+1n+1
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n?n+1?
1+2+3+…+n2n
解析:选A.∵an===,
2n+1n+1
1411
∴bn===4(-).
nn+1anan+1n?n+1?
1111111
∴Sn=4(1-+-+-+…+-)
22334nn+11
=4(1-).
n+1
6.等差数列18,15,12,…,前n项和的最大值为________. 解析:由已知得a1=18,d=-3, ∴an=a1+(n-1)d=18-3(n-1)=21-3n. ∴当n=7时,a7=0.
7×6
∴Sn最大值为S6=S7=18×7+×(-3)=63.
2
答案:63
n-1??2 ?n为正奇数?,
7.已知数列{an}中,an=?则a9=________(用数字作答),设
?2n-1 ?n为正偶数?,?
数列{an}的前n项和为Sn,则S9=________(用数字作答).
解析:a9=29-1=256.
S9=(a1+a3+a5+a7+a9)+(a2+a4+a6+a8) 1-454×?3+15?=+=377.
21-4
答案:256 377 2462n
8.数列,2,3,…,n,…,前n项的和为________.
2222
2n1
解析:由题可知,{n}的通项是等差数列{2n}的通项与等比数列{n}的通项之积.
22
2462n
设Sn=+2+3+…+n,①
2222
12462n
Sn=2+3+4+…+n1,② 22222+①-②得
1222222n(1-)Sn=+2+3+4+…+n-n1
2222222+12n=2-n1-n1
2-2+n+2
∴Sn=4-n1.
2-
n+2
答案:4-n-1 2
9.已知数列{an}是首项a1=4、公比q≠1的等比数列,Sn是其前n项和,且4a1,a5,-2a3成等差数列.
(1)求公比q的值;
(2)设An=S1+S2+S3+…+Sn,求An.
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解:(1)由已知2a5=4a1-2a3, 即2a1·q4=4a1-2a1·q2,
∵a1≠0,整理得,q4+q2-2=0, 解得q2=1,即q=1或q=-1. 又q≠1,∴q=-1.
4[1-?-1?n](2)Sn==2-2(-1)n,
1-?-1?
-1[1-?-1?n]
∴An=S1+S2+…+Sn=2n-2· 1-?-1?=2n+1-(-1)n.
11
10.(2019·日照高二检测)等差数列{an}中,a1=3,公差d=2,Sn为前n项和,求+S1S2
1
+…+. Sn
解:∵等差数列{an}的首项a1=3,公差d=2, n?n-1?n?n-1?
∴前n项和Sn=na1+d=3n+×2
22
=n2+2n(n∈N*),
111111∴=2==(-), Snn+2nn?n+2?2nn+2111∴++…+ S1S2Sn11111111111111=[(1-)+(-)+(-)+…+(-)+(-)]=(1+--) 232435nn+222n+1n+2n-1n+12n+33=-. 42?n+1??n+2?
1.在数列{an}中,a1=2,nan+1=(n+1)an+2(n∈N*),则a10等于( ) A.34 B.36 C.38 D.40 解析:选C.由nan+1=(n+1)an+2, 得(n-1)an=nan-1+2.
anan-1211则有-==2(-),
nn-1n?n-1?n-1n
an-1an-211∴-=2(-),…, n-1n-2n-2n-1a2a111an1-=2(-),累加,得-a1=2(1-). 2112nn
1
∴an=2n+2n(1-)=4n-2.∴a10=38.
n
1
2.(2019·临沂质检)已知an=n+n,则数列{an}的前n项和Sn=__________.
3111
解析:Sn=(1+2+…+n)+(+2+…+n)
333
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11=(n2+n+1-n). 23
11答案:(n2+n+1-n)
233.(2019·兖州高二检测)设数列{an}为等差数列,前n项和为Sn,已知a2=2,S5=15, (1)求{an}的通项公式;
an
(2)若bn=n,求数列{bn}的前n项和Tn.
2
???a2=2?a1+d=2
解:(1)由????a1=1,d=1,∴an=n.
???S5=15?5a1+10d=15
ann
(2)bn=n=n,
22123n
Tn=+2+3+…+n,①
22221123n
Tn=2+3+4+…+n1,② 22222+
11111n
①-②得Tn=+2+3+…+n-n1,
222222+11[1-??n]221n1nTn=-n1=1-n-n1, 2122+2+
1-21n
Tn=2-n1-n.
2-2
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