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1、考點(diǎn)規(guī)范練31 數(shù)列求和
一、基礎(chǔ)鞏固
1.數(shù)列112,314,518,7116,…,(2n-1)+12n,…的前n項(xiàng)和Sn的值等于( )
A.n2+1-12n B.2n2-n+1-12n
C.n2+1-12n-1 D.n2-n+1-12n
答案A
解析該數(shù)列的通項(xiàng)公式為an=(2n-1)+12n,則Sn=[1+3+5+…+(2n-1)]+12+122+…+12n=n2+1-12n.
2.已知數(shù)列{an}滿(mǎn)足a1=1,且對(duì)任意的n∈N*都有an+1=a1+an+n,則1an的前100項(xiàng)和為( )
A.100101 B.99100 C
2、.101100 D.200101
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)2.
∴1an=2n(n+1)=21n-1n+1.
∴1an的前100項(xiàng)和為21-12+12-13+…+1100-1101=21-1101=200101.故選D.
3.已知數(shù)列{an}滿(mǎn)足an+1-an=2,a1=-5,則|a1|+|a2|+…+|a6|=( )
A.9 B.15 C.18 D.30
答案C
解析
3、∵an+1-an=2,a1=-5,
∴數(shù)列{an}是首項(xiàng)為-5,公差為2的等差數(shù)列.
∴an=-5+2(n-1)=2n-7.
∴數(shù)列{an}的前n項(xiàng)和Sn=n(-5+2n-7)2=n2-6n.
令an=2n-7≥0,解得n≥72.
∴當(dāng)n≤3時(shí),|an|=-an;當(dāng)n≥4時(shí),|an|=an.
∴|a1|+|a2|+…+|a6|=-a1-a2-a3+a4+a5+a6
=S6-2S3=62-6×6-2(32-6×3)=18.
4.已知函數(shù)f(x)=xa的圖象過(guò)點(diǎn)(4,2),令an=1f(n+1)+f(n),n∈N*.記數(shù)列{an}的前n項(xiàng)和為Sn,則S2 018等于( )
A.
4、2018-1 B.2018+1
C.2019-1 D.2019+1
答案C
解析由f(4)=2,可得4a=2,解得a=12,則f(x)=x12.
∴an=1f(n+1)+f(n)=1n+1+n=n+1-n,
S2 018=a1+a2+a3+…+a2 018=(2-1)+(3-2)+(4-3)+…+(2019-2018)=2019-1.
5.已知數(shù)列{an}滿(mǎn)足an+1+(-1)nan=2n-1,則{an}的前60項(xiàng)和為( )
A.3 690 B.3 660 C.1 845 D.1 830
答案D
解析∵an+1+(-1)nan=2n-1,
∴當(dāng)n=2k(k∈N*)時(shí),a2
5、k+1+a2k=4k-1,①
當(dāng)n=2k+1(k∈N*)時(shí),a2k+2-a2k+1=4k+1,②
①+②得:a2k+a2k+2=8k.
則a2+a4+a6+a8+…+a60=(a2+a4)+(a6+a8)+…+(a58+a60)
=8(1+3+…+29)=8×15×(1+29)2=1800.
由②得a2k+1=a2k+2-(4k+1),
∴a1+a3+a5+…+a59=a2+a4+…+a60-[4×(0+1+2+…+29)+30]=1800-4×30×292+30=30,
∴a1+a2+…+a60=1800+30=1830.
6.已知等差數(shù)列{an},a5=π2.若函數(shù)f(x)
6、=sin 2x+1,記yn=f(an),則數(shù)列{yn}的前9項(xiàng)和為 .?
答案9
解析由題意,得yn=sin2an+1,所以數(shù)列{yn}的前9項(xiàng)和為sin2a1+sin2a2+sin2a3+…+sin2a8+sin2a9+9.
由a5=π2,得sin2a5=0.
∵a1+a9=2a5=π,∴2a1+2a9=4a5=2π,
∴2a1=2π-2a9,
∴sin2a1=sin2π-2a9=-sin2a9.
由倒序相加可得12(sin2a1+sin2a2+sin2a3+…+sin2a8+sin2a9+sin2a1+sin2a2+sin2a3+…+sin2a8+sin2a9)=0,
7、
∴y1+y2+y3+…+y8+y9=9.
7.已知數(shù)列{an}滿(mǎn)足:a3=15,an-an+1=2anan+1,則數(shù)列{anan+1}前10項(xiàng)的和為 .?
答案1021
解析∵an-an+1=2anan+1,∴an-an+1anan+1=2,即1an+1-1an=2.
∴數(shù)列1an是以2為公差的等差數(shù)列.
∵1a3=5,∴1an=5+2(n-3)=2n-1.∴an=12n-1.
∴anan+1=1(2n-1)(2n+1)=1212n-1-12n+1.
∴數(shù)列{anan+1}前10項(xiàng)的和為
121-13+13-15+…+12×10-1-12×10+1
=12×
8、1-121=12×2021=1021.
8.(2018云南昆明第二次統(tǒng)考)在數(shù)列{an}中,a1=3,{an}的前n項(xiàng)和Sn滿(mǎn)足Sn+1=an+n2.
(1)求數(shù)列{an}的通項(xiàng)公式;
(2)設(shè)數(shù)列{bn}滿(mǎn)足bn=(-1)n+2an,求數(shù)列{bn}的前n項(xiàng)和Tn.
解(1)由Sn+1=an+n2,①
得Sn+1+1=an+1+(n+1)2,②
②-①,得an=2n+1.
a1=3滿(mǎn)足上式,所以數(shù)列{an}的通項(xiàng)公式為an=2n+1.
(2)由(1)得bn=(-1)n+22n+1,
所以Tn=b1+b2+…+bn=[(-1)+(-1)2+…+(-1)n]+(23+25+…+2
9、2n+1)
=(-1)×[1-(-1)n]1-(-1)+23×(1-4n)1-4
=(-1)n-12+83(4n-1).
9.設(shè)等差數(shù)列{an}的公差為d,前n項(xiàng)和為Sn,等比數(shù)列{bn}的公比為q,已知b1=a1,b2=2,q=d,S10=100.
(1)求數(shù)列{an},{bn}的通項(xiàng)公式;
(2)當(dāng)d>1時(shí),記cn=anbn,求數(shù)列{cn}的前n項(xiàng)和Tn.
解(1)由題意,得10a1+45d=100,a1d=2,即2a1+9d=20,a1d=2,
解得a1=1,d=2或a1=9,d=29.
故an=2n-1,bn=2n-1或an=19(2n+79),bn=9·29n-1.
10、
(2)由d>1,知an=2n-1,bn=2n-1,故cn=2n-12n-1,
于是Tn=1+32+522+723+924+…+2n-12n-1,①
12Tn=12+322+523+724+925+…+2n-12n.②
①-②可得12Tn=2+12+122+…+12n-2-2n-12n=3-2n+32n,故Tn=6-2n+32n-1.
10.已知Sn為數(shù)列{an}的前n項(xiàng)和,an>0,an2+2an=4Sn+3.
(1)求{an}的通項(xiàng)公式;
(2)設(shè)bn=1anan+1,求數(shù)列{bn}的前n項(xiàng)和.
解(1)由an2+2an=4Sn+3,可知an+12+2an+1=4Sn+1+3
11、.
兩式相減可得an+12-an2+2(an+1-an)=4an+1,
即2(an+1+an)=an+12-an2=(an+1+an)·(an+1-an).
由于an>0,可得an+1-an=2.
又a12+2a1=4a1+3,
解得a1=-1(舍去),a1=3.
所以{an}是首項(xiàng)為3,公差為2的等差數(shù)列,故{an}的通項(xiàng)公式為an=2n+1.
(2)由an=2n+1可知
bn=1anan+1=1(2n+1)(2n+3)=1212n+1-12n+3.
設(shè)數(shù)列{bn}的前n項(xiàng)和為T(mén)n,
則Tn=b1+b2+…+bn
=1213-15+15-17+…+12n+1-12n+3
12、
=n3(2n+3).
11.已知各項(xiàng)均為正數(shù)的數(shù)列{an}的前n項(xiàng)和為Sn,滿(mǎn)足an+12=2Sn+n+4,a2-1,a3,a7恰為等比數(shù)列{bn}的前3項(xiàng).
(1)求數(shù)列{an},{bn}的通項(xiàng)公式;
(2)若cn=(-1)nlog2bn-1anan+1,求數(shù)列{cn}的前n項(xiàng)和Tn.
解(1)因?yàn)閍n+12=2Sn+n+4,
所以an2=2Sn-1+n-1+4(n≥2).
兩式相減,得an+12-an2=2an+1,
所以an+12=an2+2an+1=(an+1)2.
因?yàn)閧an}是各項(xiàng)均為正數(shù)的數(shù)列,
所以an+1=an+1,即an+1-an=1.
又a32=(
13、a2-1)a7,所以(a2+1)2=(a2-1)(a2+5),
解得a2=3,a1=2,
所以{an}是以2為首項(xiàng),1為公差的等差數(shù)列,
所以an=n+1.
由題意知b1=2,b2=4,b3=8,故bn=2n.
(2)由(1)得cn=(-1)nlog22n-1(n+1)(n+2)
=(-1)nn-1(n+1)(n+2),
故Tn=c1+c2+…+cn=[-1+2-3+…+(-1)nn]-12×3+13×4+…+1(n+1)×(n+2).
設(shè)Fn=-1+2-3+…+(-1)nn,
則當(dāng)n為偶數(shù)時(shí),Fn=(-1+2)+(-3+4)+…+[-(n-1)+n]=n2;
當(dāng)n為奇數(shù)時(shí)
14、,Fn=Fn-1+(-n)=n-12-n=-(n+1)2.
設(shè)Gn=12×3+13×4+…+1(n+1)×(n+2),
則Gn=12-13+13-14+…+1n+1-1n+2=12-1n+2.
所以Tn=n-12+1n+2,n為偶數(shù),-n+22+1n+2,n為奇數(shù).
二、能力提升
12.在數(shù)列{an}中,a1=1,且an+1=an2an+1.若bn=anan+1,則數(shù)列{bn}的前n項(xiàng)和Sn為( )
A.2n2n+1 B.n2n+1 C.2n2n-1 D.2n-12n+1
答案B
解析由an+1=an2an+1,得1an+1=1an+2,
∴數(shù)列1an是以1為首項(xiàng),2為公差
15、的等差數(shù)列,
∴1an=2n-1,又bn=anan+1,
∴bn=1(2n-1)(2n+1)=1212n-1-12n+1,
∴Sn=1211-13+13-15+…+12n-1-12n+1=n2n+1,故選B.
13.(2018福建寧德期末)今要在一個(gè)圓周上標(biāo)出一些數(shù),第一次先把圓周二等分,在這兩個(gè)分點(diǎn)處分別標(biāo)上1,如圖(1)所示;第二次把兩段半圓弧二等分,在這兩個(gè)分點(diǎn)處分別標(biāo)上2,如圖(2)所示;第三次把四段圓弧二等分,并在這4個(gè)分點(diǎn)處分別標(biāo)上3,如圖(3)所示.如此繼續(xù)下去,當(dāng)?shù)趎次標(biāo)完數(shù)以后,這個(gè)圓周上所有已標(biāo)出的數(shù)的總和是 .?
答案(n-1)×2n+2
16、
解析由題意可得,第n次標(biāo)完后,圓周上所有已標(biāo)出的數(shù)的總和為T(mén)n=1+1+2×2+3×22+…+n×2n-1.
設(shè)S=1+2×2+3×22+…+n×2n-1,
則2S=2+2×22+…+(n-1)×2n-1+n×2n,
兩式相減可得-S=1+2+22+…+2n-1-n×2n=(1-n)×2n-1,則S=(n-1)×2n+1,故Tn=(n-1)×2n+2.
14.已知首項(xiàng)為32的等比數(shù)列{an}不是遞減數(shù)列,其前n項(xiàng)和為Sn(n∈N*),且S3+a3,S5+a5,S4+a4成等差數(shù)列.
(1)求數(shù)列{an}的通項(xiàng)公式;
(2)設(shè)bn=(-1)n+1·n(n∈N*),求數(shù)列{an·bn
17、}的前n項(xiàng)和Tn.
解(1)設(shè)等比數(shù)列{an}的公比為q.
由S3+a3,S5+a5,S4+a4成等差數(shù)列,
可得2(S5+a5)=S3+a3+S4+a4,
即2(S3+a4+2a5)=2S3+a3+2a4,
即4a5=a3,則q2=a5a3=14,解得q=±12.
由等比數(shù)列{an}不是遞減數(shù)列,可得q=-12,
故an=32·-12n-1=(-1)n-1·32n.
(2)由bn=(-1)n+1·n,可得an·bn=(-1)n-1·32n·(-1)n+1·n=3n·12n.
故前n項(xiàng)和Tn=31·12+2·122+…+n·12n,
則12Tn=31·122+2·123+…
18、+n·12n+1,
兩式相減可得,12Tn=312+122+…+12n-n·
12n+1=3121-12n1-12-n·12n+1,
化簡(jiǎn)可得Tn=61-n+22n+1.
15.(2018湖南長(zhǎng)沙雅禮中學(xué)模擬)若數(shù)列{an}的前n項(xiàng)和Sn滿(mǎn)足Sn=2an-λ(λ>0,n∈N*).
(1)證明:數(shù)列{an}為等比數(shù)列,并求an;
(2)若λ=4,bn=an,n是奇數(shù),log2an,n是偶數(shù)(n∈N*),求數(shù)列{bn}的前2n項(xiàng)和T2n.
(1)證明∵Sn=2an-λ,
當(dāng)n=1時(shí),得a1=λ,
當(dāng)n≥2時(shí),Sn-1=2an-1-λ,
則Sn-Sn-1=2an-2an-1,
19、即an=2an-2an-1,∴an=2an-1,
∴數(shù)列{an}是以λ為首項(xiàng),2為公比的等比數(shù)列,
∴an=λ·2n-1.
(2)解∵λ=4,∴an=4·2n-1=2n+1,
∴bn=2n+1,n是奇數(shù),n+1,n是偶數(shù).
∴T2n=22+3+24+5+26+7+…+22n+2n+1
=(22+24+26+…+22n)+(3+5+…+2n+1)
=4-22n·41-4+n(3+2n+1)2=4n+1-43+n(n+2),
∴T2n=4n+13+n2+2n-43.
三、高考預(yù)測(cè)
16.已知數(shù)列{an}的前n項(xiàng)和為Sn,且a1=2,Sn=2an+k,等差數(shù)列{bn}的前n項(xiàng)和為
20、Tn,且Tn=n2.
(1)求k和Sn;
(2)若cn=an·bn,求數(shù)列{cn}的前n項(xiàng)和Mn.
解(1)∵Sn=2an+k,∴當(dāng)n=1時(shí),S1=2a1+k.
∴a1=-k=2,即k=-2.∴Sn=2an-2.
∴當(dāng)n≥2時(shí),Sn-1=2an-1-2.
∴an=Sn-Sn-1=2an-2an-1.∴an=2an-1.
∴數(shù)列{an}是以2為首項(xiàng),2為公比的等比數(shù)列.即an=2n.
∴Sn=2n+1-2.
(2)∵等差數(shù)列{bn}的前n項(xiàng)和為T(mén)n,且Tn=n2,
∴當(dāng)n≥2時(shí),bn=Tn-Tn-1=2n-1.
又b1=T1=1符合bn=2n-1,
∴bn=2n-1.
∴cn=an·bn=(2n-1)2n.
∴數(shù)列{cn}的前n項(xiàng)和Mn=1×2+3×22+5×23+…+(2n-3)×2n-1+(2n-1)×2n,①
∴2Mn=1×22+3×23+5×24+…+(2n-3)×2n+(2n-1)×2n+1,②
由①-②,得-Mn=2+2×22+2×23+2×24+…+2×2n-(2n-1)×2n+1=2+2×22-2n+11-2-(2n-1)×2n+1,
即Mn=6+(2n-3)2n+1.
10
廣西2020版高考數(shù)學(xué)一輪復(fù)習(xí) 考點(diǎn)規(guī)范練31 數(shù)列求和 文
