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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.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}滿足an+1-an=2,a1=-5,則|a1|+|a2|+…+|a6|=(　　) A.9 B.15 C.18 D.30 答案C 解析∵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=62-66-2(32-63)=18. 4.已知函數(shù)f(x)=xa的圖象過點(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.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),a2k+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)=815(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]=1800-430292+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)=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, ∴y1+y2+y3+…+y8+y9=9. 7.已知數(shù)列{an}滿足: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+…+1210-1-1210+1 =121-121=122021=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+…+22n+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=929n-1. (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. 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Tn, 則Tn=b1+b2+…+bn =1213-15+15-17+…+12n+1-12n+3 =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=(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]-123+134+…+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),Fn=Fn-1+(-n)=n-12-n=-(n+1)2. 設(shè)Gn=123+134+…+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 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Tn=1+1+22+322+…+n2n-1. 設(shè)S=1+22+322+…+n2n-1, 則2S=2+222+…+(n-1)2n-1+n2n, 兩式相減可得-S=1+2+22+…+2n-1-n2n=(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+1n(n∈N*),求數(shù)列{anbn}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-132n. (2)由bn=(-1)n+1n,可得anbn=(-1)n-132n(-1)n+1n=3n12n. 故前n項(xiàng)和Tn=3112+2122+…+n12n, 則12Tn=31122+2123+…+n12n+1, 兩式相減可得,12Tn=312+122+…+12n-n 12n+1=3121-12n1-12-n12n+1, 化簡可得Tn=61-n+22n+1. 15.(2018湖南長沙雅禮中學(xué)模擬)若數(shù)列{an}的前n項(xiàng)和Sn滿足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, 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=λ2n-1. (2)解∵λ=4,∴an=42n-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) =4-22n41-4+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Tn,且Tn=n2. (1)求k和Sn; (2)若cn=anbn,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Tn,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=anbn=(2n-1)2n. ∴數(shù)列{cn}的前n項(xiàng)和Mn=12+322+523+…+(2n-3)2n-1+(2n-1)2n, ① ∴2Mn=122+323+524+…+(2n-3)2n+(2n-1)2n+1, ② 由①-②,得-Mn=2+222+223+224+…+22n-(2n-1)2n+1=2+222-2n+11-2-(2n-1)2n+1, 即Mn=6+(2n-3)2n+1.