习题课:数列求和1.已知数列{an}的通项公式an=,其前n项和Sn=,则项数n等于()A.13B.10C.9D.6解析:an==1-.∴Sn=n-=n-1+=5+,∴n=6.答案:D2.若数列{an}的通项公式是an=(-1)n(3n-2),则a1+a2+…+a10等于()A.15B.12C.-12D.-15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.答案:A3.数列1,1+2,1+2+22,…,1+2+22+…+2n-1,…的前n项和为()A.2n-n-1B.2n+1-n-2C.2nD.2n+1-n解析:∵an=2n-1,∴Sn=(2+22+…+2n)-n=2n+1-n-2.答案:B4.若数列{an}为等比数列,且a1=1,q=2,则Tn=+…+的结果可化为()A.1-B.1-C.D.解析:∵an=2n-1,∴.1∴数列是以为首项,以为公比的等比数列,所以其前n项和Tn=.答案:C5.已知数列{an}:,…,那么数列{bn}=前n项的和为()A.4B.4C.1-D.解析:∵an=,∴bn==4.∴Sn=4=4.答案:A6.在等比数列{an}中,若a1=,a4=-4,则|a1|+|a2|+|a3|+…+|an|=.解析:设{an}的公比为q.∵{an}为等比数列,且a1=,a4=-4,∴q3==-8,∴q=-2.∴an=×(-2)n-1,∴|an|=2n-2.∴|a1|+|a2|+|a3|+…+|an|2=.答案:7.数列11,103,1005,10007,…的前n项和Sn=.解析:数列的通项公式an=10n+(2n-1).所以Sn=(10+1)+(102+3)+…+(10n+2n-1)=(10+102+…+10n)+[1+3+…+(2n-1)]=(10n-1)+n2.答案:(10n-1)+n28.数列…的前n项和等于.解析:∵an=,∴Sn===.答案:9.已知等差数列{an}的前n项和Sn满足S3=0,S5=-5.(1)求{an}的通项公式;(2)求数列的前n项和.解:(1)设{an}的公差为d,则Sn=na1+d.3由已知可得解得a1=1,d=-1.故{an}的通项公式为an=2-n.(2)由(1)知=,从而数列的前n项和为=.10.已知数列{an}的前n项和为Sn,且Sn=2n2+n,n∈N*,数列{bn}满足an=4log2bn+3,n∈N*.(1)求an,bn;(2)求数列{anbn}的前n项和Tn.解:(1)由Sn=2n2+n,得当n=1时,a1=S1=3;当n≥2时,an=Sn-Sn-1=4n-1.所以an=4n-1,n∈N*.由4n-1=an=4log2bn+3,得bn=2n-1,n∈N*.(2)由(1)知anbn=(4n-1)·2n-1,n∈N*.所以Tn=3+7×2+11×22+…+(4n-1)·2n-1,2Tn=3×2+7×22+…+(4n-5)·2n-1+(4n-1)·2n,所以2Tn-Tn=(4n-1)2n-[3+4(2+22+…+2n-1)]=(4n-5)2n+5.故Tn=(4n-5)2n+5,n∈N*.4
