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+12nC.\(n+1\)n2D.n2+n+2n+1答案:A解析: an=11+2+3+…+n=2n\(n+1\)=2(1n-1n+1),∴Sn=a1+a2+…+an=2[(1-12)+(12-13)+…+(1n-1n+1)]=2(1-1n+1)=2nn+1.5.11×3+13×5+15×7+…+1\(2n-1\)\(2n+1\)=.答案:n2n+1解析: 1\(2n-1\)\(2n+1\)=12(12n-1-12n+1),∴11×3+13×5+15×7+…+1\(2n-1\)\(2n+1\)=12(1-13+13-15+15-17+…+12n-1-12n+1)=12(1-12n+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=12,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\(3+3n\)=23(1n-1n+1),∴Tn=23(1-12+12-13+…+1n-1n+1)=23(1-1n+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.答案:4-n+22n-1解析:设Sn=22+422+623+…+2n2n,①12Sn=222+423+624+…+2n2n+1,②①-②得(1-12)Sn=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...
