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1、经典小初高讲义课时训练14 数列求和一、分组求和1.若数列{an}的通项公式是an=(-1)n(3n-2),则a1+a2+…+a10=( ) A.15B.12C.-12D.-15答案:A解析:∵an=(-1)n(3n-2),则a1+a2+…+a10=-1+4-7+10-…-25+28=(-1+4)+(-7+10)+…+(-25+28)=3×5=15.2.已知数列{an}满足a1=1,an+1=an+n+2n(n∈N*),则an为( )A.n(n-1)2+2n-1-1B.n(n-1)2+2n-1C.n(n+1)2+2n+1-1D.n(n-1)2+2n+1-1答
2、案:B解析:∵an+1=an+n+2n,∴an+1-an=n+2n.∴an=a1+(a2-a1)+(a3-a2)+…+(an-an-1)=1+(1+2)+(2+22)+…+[(n-1)+2n-1]=1+[1+2+3+…+(n-1)]+(2+22+…+2n-1)=1+(n-1)n2+2(1-2n-1)1-2=n(n-1)2+2n-1.3.(2015广东湛江高二期末,19)已知数列{an}为等差数列,a5=5,d=1;数列{bn}为等比数列,b4=16,q=2.(1)求数列{an},{bn}的通项公式an,bn;(2)设cn=an+bn,求数列{cn}的前n项和Tn.解:(1)∵数列{an}为等
3、差数列,a5=5,d=1,∴a1+4=5,解得a1=1,∴an=1+(n-1)×1=n.∵数列{bn}为等比数列,b4=16,q=2,∴b1·23=16,解得b1=2,∴bn=2×2n-1=2n.(2)∵cn=an+bn=n+2n,∴Tn=(1+2+3+…+n)+(2+22+23+…+2n)=n(n+1)2+2(1-2n)1-2=n2+n2+2n+1-2.二、裂项相消法求和4.数列{an}的通项公式an=11+2+3+…+n,则其前n项和Sn=( )A.2nn+1B.n+12n小初高优秀教案经典小初高讲义C.(n+1)n2D.n2+n+2n+1答案:A解析:∵an=11+2+3+…+n=2
4、n(n+1)=21n-1n+1,∴Sn=a1+a2+…+an=21-12+12-13+…+1n-1n+1=21-1n+1=2nn+1.5.11×3+13×5+15×7+…+1(2n-1)(2n+1)= . 答案:n2n+1解析:∵1(2n-1)(2n+1)=1212n-1-12n+1,∴11×3+13×5+15×7+…+1(2n-1)(2n+1)=121-13+13-15+15-17+…+12n-1-12n+1=121-12n+1=n2n+1.6.(2015山东省潍坊四县联考,17)等差数列{an}中,a1=3,其前n项和为Sn.等比数列{bn}的各项均为正数,b1=1,且b2+S2=1
5、2,a3=b3.(1)求数列{an}与{bn}的通项公式;(2)求数列1Sn的前n项和Tn.解:(1)设数列{an}的公差为d,数列{bn}的公比为q,由已知可得q+3+3+d=12,q2=3+2d,又q>0,∴d=3,q=3,∴an=3+3(n-1)=3n,bn=3n-1.(2)由(1)知数列{an}中,a1=3,an=3n,∴Sn=n(3+3n)2,∴1Sn=2n(3+3n)=231n-1n+1,∴Tn=231-12+12-13+…+1n-1n+1=231-1n+1=2n3(n+1).三、错位相减法求和7.数列22,422,623,…,2n2n,…前n项的和为 . 小初高优秀教案经
6、典小初高讲义答案:4-n+22n-1解析:设Sn=22+422+623+…+2n2n,①12Sn=222+423+624+…+2n2n+1,②①-②得1-12Sn=22+222+223+224+…+22n-2n2n+1=2-12n-1-2n2n+1.∴Sn=4-n+22n-1.8.(2015湖北高考,文19)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)由题意有,10a1+45d=100,a1d=2
7、,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.(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
