一、非標(biāo)準(zhǔn)
    1.若數(shù)列{an}的首項(xiàng)a1=1,且an=an-1+2(n≥2),則a7等于(　　)
    A.13 B.14 C.15 D.17
    2.已知Sn為等差數(shù)列{an}的前n項(xiàng)和,a2+a8=6,則S9等于(　　)
    A. B.27 C.54 D.108
    3.在等差數(shù)列{an}中,a2=3,a3+a4=9,則a1a6的值為(　　)
    A.14 B.18 C.21 D.27
    4.在等差數(shù)列{an}中,a5+a6+a7=15,那么a3+a4+…+a9等于(　　)
    A.21 B.30 C.35 D.40
    5.(2014天津河西口模擬)設(shè)等差數(shù)列{an}的前n項(xiàng)和為Sn,若a11-a8=3,S11-S8=3,則使an>0的最小正整數(shù)n的值是(　　)
    A.8 B.9 C.10 D.11
    6.(2014浙江聯(lián)考)已知每項(xiàng)均大于零的數(shù)列{an}中,首項(xiàng)a1=1,且前n項(xiàng)和Sn滿足Sn-Sn-1=2(nN+,且n≥2),則a81等于(　　)
    A.638 B.639 C.640 D.641
    7.若等差數(shù)列{an}滿足a7+a8+a9>0,a7+a10<0,則當(dāng)n=　　　　　時(shí),{an}的前n項(xiàng)和.
    8.若等差數(shù)列{an}前9項(xiàng)的和等于前4項(xiàng)的和,且ak+a4=0,則k=　　　　　.
    9.已知公差大于零的等差數(shù)列{an}的前n項(xiàng)和為Sn,且滿足a3·a4=117,a2+a5=22.
    (1)求數(shù)列{an}的通項(xiàng)公式;
    (2)若數(shù)列{bn}滿足bn=,是否存在非零實(shí)數(shù)c使得{bn}為等差數(shù)列?若存在,求出c的值;若不存在,請(qǐng)說明理由.
    10.已知數(shù)列{an}的前n項(xiàng)和為Sn,a1=1,an≠0,anan+1=λSn-1,其中λ為常數(shù).
    (1)證明:an+2-an=λ;
    (2)是否存在λ,使得{an}為等差數(shù)列?并說明理由.
    11.(2014遼寧,文9)設(shè)等差數(shù)列{an}的公差為d.若數(shù)列{}為遞減數(shù)列,則(　　)
    A.d>0 B.d<0 C.a1d>0 D.a1d<0
    12.已知等差數(shù)列{an}的前n項(xiàng)和為Sn,S4=40,Sn=210,Sn-4=130,則n等于(　　)
    A.12 B.14 C.16 D.18
    13.若數(shù)列{an}滿足:a1=19,an+1=an-3(nN+),則數(shù)列{an}的前n項(xiàng)和數(shù)值時(shí),n的值為(　　)
    A.6 B.7 C.8 D.9
    14.已知正項(xiàng)數(shù)列{an}滿足:a1=1,a2=2,2(nN+,n≥2),則a7=　　　　　.
    15.已知數(shù)列{an}的各項(xiàng)均為正數(shù),前n項(xiàng)和為Sn,且滿足2Sn=+n-4(nN+).
    (1)求證:數(shù)列{an}為等差數(shù)列;
    (2)求數(shù)列{an}的通項(xiàng)公式.
    16.設(shè)數(shù)列{an}的前n項(xiàng)和為Sn,a1=1,an=+2(n-1)(nN+).
    (1)求證:數(shù)列{an}為等差數(shù)列,并求an與Sn;
    (2)是否存在自然數(shù)n,使得S1++…+-(n-1)2=2015?若存在,求出n的值;若不存在,請(qǐng)說明理由.
    一、非標(biāo)準(zhǔn)
    1.A　解析:an=an-1+2(n≥2),
    ∴an-an-1=2.
    又a1=1,∴數(shù)列{an}是以1為首項(xiàng),以2為公差的等差數(shù)列,
    故a7=1+2×(7-1)=13.
    2.B　解析:S9==27.
    3.A　解析:設(shè)等差數(shù)列{an}的公差為d,
    則依題意得由此解得
    所以a6=a1+5d=7,a1a6=14.
    4.C　解析:由題意得3a6=15,a6=5.
    所以a3+a4+…+a9=7a6=7×5=35.
    5.C　解析:設(shè)等差數(shù)列{an}的公差為d,
    a11-a8=3d=3,∴d=1.
    ∵S11-S8=a11+a10+a9=3a1+27d=3,
    ∴a1=-8,∴令an=-8+(n-1)>0,解得n>9.
    因此使an>0的最小正整數(shù)n的值是10.
    6.C　解析:由已知Sn-Sn-1=2,可得=2,
    {}是以1為首項(xiàng),2為公差的等差數(shù)列,
    故=2n-1,Sn=(2n-1)2,
    a81=S81-S80=1612-1592=640,故選C.
    7.8　解析:由等差數(shù)列的性質(zhì)可得a7+a8+a9=3a8>0,即a8>0;而a7+a10=a8+a9<0,故a9<0.所以數(shù)列{an}的前8項(xiàng)和.
    8.10　解析:設(shè)等差數(shù)列{an}的前n項(xiàng)和為Sn,則S9-S4=0,
    即a5+a6+a7+a8+a9=0,5a7=0,故a7=0.
    而ak+a4=0=2a7,故k=10.
    9.解:(1)設(shè)等差數(shù)列{an}的公差為d,且d>0,
    由等差數(shù)列的性質(zhì),得a2+a5=a3+a4=22,
    所以a3,a4是關(guān)于x的方程x2-22x+117=0的解,
    所以a3=9,a4=13.
    易知a1=1,d=4,故所求通項(xiàng)為an=1+(n-1)×4=4n-3.
    (2)由(1)知Sn==2n2-n,
    所以bn=.
    (方法一)所以b1=,b2=,b3=(c≠0).
    令2b2=b1+b3,解得c=-.
    當(dāng)c=-時(shí),bn==2n,
    當(dāng)n≥2時(shí),bn-bn-1=2.
    故當(dāng)c=-時(shí),數(shù)列{bn}為等差數(shù)列.
    (方法二)bn=.
    c≠0,∴可令c=-,得到bn=2n.
    bn+1-bn=2(n+1)-2n=2(n∈N+),
    ∴數(shù)列{bn}是公差為2的等差數(shù)列.
    故存在一個(gè)非零常數(shù)c=-,使數(shù)列{bn}也為等差數(shù)列.
    10.解:(1)由題設(shè),anan+1=λSn-1,an+1an+2=λSn+1-1,
    兩式相減,得an+1(an+2-an)=λan+1.
    由于an+1≠0,所以an+2-an=λ.
    (2)由題設(shè),a1=1,a1a2=λS1-1,可得a2=λ-1.
    由(1)知,a3=λ+1.
    令2a2=a1+a3,解得λ=4.
    故an+2-an=4.
    由此可得{a2n-1}是首項(xiàng)為1,公差為4的等差數(shù)列,a2n-1=4n-3;{a2n}是首項(xiàng)為3,公差為4的等差數(shù)列,a2n=4n-1.
    所以an=2n-1,an+1-an=2.
    因此存在λ=4,使得數(shù)列{an}為等差數(shù)列.
    11.D　解析:{}為遞減數(shù)列,
    =<1.
    ∴a1d<0.故選D.
    12.B　解析:易得Sn-Sn-4=an+an-1+an-2+an-3=80.
    又S4=a1+a2+a3+a4=40,
    所以4(a1+an)=120,a1+an=30.
    由Sn==210,得n=14.
    13.B　解析:a1=19,an+1-an=-3,
    ∴數(shù)列{an}是以19為首項(xiàng),-3為公差的等差數(shù)列.
    an=19+(n-1)×(-3)=22-3n.
    設(shè){an}的前k項(xiàng)和數(shù)值,
    則有kN+.
    ∴
    ∴≤k≤.
    ∵k∈N+,∴k=7.
    ∴滿足條件的n的值為7.
    14.　解析:因?yàn)?(nN+,n≥2),
    所以數(shù)列{}是以=1為首項(xiàng),以d==4-1=3為公差的等差數(shù)列.
    所以=1+3(n-1)=3n-2.
    所以an=,n≥1.
    所以a7=.
    15.(1)證明:當(dāng)n=1時(shí),有2a1=+1-4,即-2a1-3=0,
    解得a1=3(a1=-1舍去).
    當(dāng)n≥2時(shí),有2Sn-1=+n-5.
    又2Sn=+n-4,
    兩式相減得2an=+1,
    即-2an+1=,
    也即(an-1)2=,因此an-1=an-1或an-1=-an-1.
    若an-1=-an-1,則an+an-1=1.
    而a1=3,所以a2=-2,這與數(shù)列{an}的各項(xiàng)均為正數(shù)相矛盾,
    所以an-1=an-1,即an-an-1=1.
    因此,數(shù)列{an}為首項(xiàng)為3,公差為1的等差數(shù)列.
    (2)解:由(1)知a1=3,d=1,所以數(shù)列{an}的通項(xiàng)公式an=3+(n-1)×1=n+2,即an=n+2.
    16.(1)證明:由an=+2(n-1),得Sn=nan-2n(n-1)(nN+).
    當(dāng)n≥2時(shí),an=Sn-Sn-1=nan-(n-1)an-1-4(n-1),
    即an-an-1=4,
    故數(shù)列{an}是以1為首項(xiàng),4為公差的等差數(shù)列.
    于是,an=4n-3,Sn==2n2-n(nN+).
    (2)解:由(1),得=2n-1(nN+).
    又S1++…+-(n-1)2=1+3+5+7+…+(2n-1)-(n-1)2=n2-(n-1)2=2n-1.
    令2n-1=2015,得n=1008,
    即存在滿足條件的自然數(shù)n=1008.