考点规范练31数列求和一、基础巩固1.数列112,314,518,7116,…,(2n-1)+12n,…的前n项和Sn的值等于()A.n2+1-12nB.2n2-n+1-12nC.n2+1-12n-1D.n2-n+1-12n答案A解析该数列的通项公式为an=(2n-1)+12n,则Sn=[1+3+5+…+(2n-1)]+(12+122+…+12n)=n2+1-12n.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{1an}的前100项和为()A.100101B.99100C.101100D.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)=2(1n-1n+1).∴{1an}的前100项和为2(1-12+12-13+…+1100-1101)=2(1-1101)=200101.故选D.3.已知数列{an}满足an+1-an=2,a1=-5,则|a1|+|a2|+…+|a6|=()A.9B.15C.18D.30答案C解析 an+1-an=2,a1=-5,∴数列{an}是首项为-5,公差为2的等差数列.∴an=-5+2(n-1)=2n-7.∴数列{an}的前n项和Sn=n(-5+2n-7)2=n2-6n.令an=2n-7≥0,解得n≥72.∴当n≤3时,|an|=-an;当n≥4时,|an|=an.∴|a1|+|a2|+…+|a6|=-a1-a2-a3+a4+a5+a6=S6-2S3=62-6×6-2(32-6×3)=18.4.已知函数f(x)=xa的图象过点(4,2),令an=1f(n+1)+f(n),n∈N*.记数列{an}的前n项和为Sn,则S2018等于()A.❑√2018-1B.❑√2018+1C.❑√2019-1D.❑√2019+1答案C解析由f(4)=2,可得4a=2,解得a=12,则f(x)=x12.∴an=1f(n+1)+f(n)=1❑√n+1+❑√n=❑√n+1−❑√n,S2018=a1+a2+a3+…+a2018=(❑√2−❑√1)+(❑√3−❑√2)+(❑√4−❑√3)+…+(❑√2019−❑√2018)=❑√2019-1.5.已知数列{an}满足an+1+(-1)nan=2n-1,则{an}的前60项和为()A.3690B.3660C.1845D.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+2-a2k+1=4k+1,②①+②得:a2k+a2k+2=8k.则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.已知等差数列{an},a5=π2.若函数f(x)=sin2x+1,记yn=f(an),则数列{yn}的前9项和为.答案9解析由题意,得yn=sin2an+1,所以数列{yn}的前9项和为sin2a1+sin2a2+sin2a3+…+sin2a8+sin2a9+9.由a5=π2,得sin2a5=0. a1+a9=2a5=π,∴2a1+2a9=4a5=2π,∴2a1=2π-2a9,∴sin2a1=sin(2π-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.已知数列{an}满足:a3=15,an-an+1=2anan+1,则数列{anan+1}前10项的和为.答案1021解析 an-an+1=2anan+1,∴an-an+1anan+1=2,即1an+1−1an=2.∴数列{1an}是以2为公差的等差数列. 1a3=5,∴1an=5+2(n-3)=2n-1.∴an=12n-1.∴anan+1=1(2n-1)(2n+1)=12(12n-1-12n+1).∴数列{anan+1}前10项的和为12(1-13+13-15+…+12×10-1-12×10+1)=12×(1-121)=12×2021=1021.8.(2018云南昆明第二次统考)在数列{an}中,a1=3,{an}的前n项和Sn满足Sn+1=an+n2.(1)求数列{an}的通项公式;(2)设数列{bn}满足bn=(-1)n+2an,求数列{bn}的前n项和Tn.解(1)由Sn+1=an+n2,①得Sn+1+1=an+1+(n+1)2,②②-①,得an=2n+1.a1=3满足上式,所以数列{an}的通项公式为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.设等差数列{an}的公差为d,前n项和为Sn,等比数列{bn}的公比为q,已知b1=a1,b2=2,q=d,S10=100.(1)求数列{an},{bn}的通项公式;(2)当d>1时,记cn=anbn,求数列{cn}的前n项和Tn.解(1)由题意,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·(29)n-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-1,①12Tn=12+322+523+724+925+…+2n-12n.②①-②可得12Tn=2+12+122+…+12n-2−2n-12n=3-2n+32n,故Tn=6-2n+32n-1.10.已知Sn为数列{an}的前n项和,an>0,an2+2an=4Sn+3.(1)求{an}的通项公式;(2)设bn=1anan+1,求数列{bn}的前n项和.解(1)由an2+2an=4...
