高中數(shù)學(xué)第四章數(shù)列4.2.2等差數(shù)列的前n項和公式第1課時等差數(shù)列的前n項和公式訓(xùn)練提升新人教版選修2

4.2.2等差數(shù)列的前n項和公式第1課時等差數(shù)列的前n項和公式課后·訓(xùn)練提升基礎(chǔ)鞏固1.在等差數(shù)列{an}中,若a2=1,a4=5,則數(shù)列{an}的前5項和S5等于()A.7 B.15 C.20 D.25答案:B解析:設(shè)等差數(shù)列{an}的公差為d,則有a故S5=5a1+5×422.已知等差數(shù)列{an}的前n項和Sn=n2+5n,則公差d等于()A.1 B.2 C.5 D.10答案:B解析:∵a1=S1=6,a1+a2=S2=14,∴a2=8,∴d=a2-a1=2.3.已知{an}是等差數(shù)列,a1=10,前10項和S10=70,則其公差d等于()A.-23 B.-13 C.13答案:A解析:因為S10=10a1+10×92d=70,a所以d=-234.已知數(shù)列{an}滿足2an=an-1+an+1(n>1),Sn是數(shù)列{an}的前n項和,a2,a2021是函數(shù)f(x)=x2-6x+5的兩個零點,則S2022的值為()A.6066 B.12 C.2020 D.6060答案:A解析:∵數(shù)列{an}滿足2an=an-1+an+1(n>1),∴數(shù)列{an}為等差數(shù)列,又a2,a2021是函數(shù)f(x)=x2-6x+5的兩個零點,即a2,a2021是方程x2-6x+5=0的兩個根,∴a2+a2021=6,∴S2022=(a1+5.已知{an}是公差d=1的等差數(shù)列,Sn為數(shù)列{an}的前n項和,若S8=4S4,則a10等于()A.172 B.192 C.10 D答案:B解析:由題意,可知S8=8a1+8×72d=8a1+28,S4=4a1+4×32d=4a1+6.因為S8=4S4,即8a1+28=4(4a1+6),解得a1=12,所以a106.在等差數(shù)列{an}和{bn}中,若a1+b100=100,b1+a100=100,則數(shù)列{an+bn}的前100項和為()A.0 B.100 C.1000 D.10000答案:D解析:數(shù)列{an+bn}的前100項和為100(a1+a100)2+100(=50×200=10000.7.在等差數(shù)列{an}中,如果a1+a2+a3=-24,a18+a19+a20=78,那么此數(shù)列的前20項和為()A.160 B.180 C.200 D.220答案:B解析:由題意可知a1+a2+a3=3a2=-24,得a2=-8,由a18+a19+a20=3a19=78,得a19=26,于是S20=10(a1+a20)=10(a2+a19)=10×(-8+26)=180.8.已知{an}是等差數(shù)列,Sn是它的前n項和.若S4=20,a4=8,則S8=.

答案:72解析:設(shè)等差數(shù)列{an}的公差為d,則由S4=4a1+故S8=8×2+8×72×29.已知數(shù)列{an}滿足a1=50,an+1=an+2n.(1)求{an}的通項公式;(2)已知數(shù)列{bn}的前n項和為an,若bm=50,求正整數(shù)m的值.解:(1)當n≥2時,an=(an-an-1)+(an-1-an-2)+…+(a3-a2)+(a2-a1)+a1=2(n-1)+2(n-2)+…+2×2+2×1+50=2×(n-1)n2又a1=50=12-1+50,故{an}的通項公式為an=n2-n+50.(2)b1=a1=50,當n≥2時,bn=an-an-1=n2-n+50-[(n-1)2-(n-1)+50]=2n-2,即bn=50當m≥2時,令bm=50,得2m-2=50,解得m=26.又b1=50,故正整數(shù)m的值為1或26.10.已知等差數(shù)列{an}的前三項依次為a,4,3a,前n項和為Sn,且Sk=110.(1)求a及k的值;(2)設(shè)數(shù)列{bn}的通項公式為bn=Snn,證明:數(shù)列{bn}是等差數(shù)列,并求其前n項和T(1)解:設(shè)等差數(shù)列{an}的公差為d,則a1=a,a2=4,a3=3a,由已知有a+3a=8,得a1=a=2,d=4-2=2,故Sk=ka1+k(k-1)2·d=2k+k由Sk=110,得k2+k-110=0,解得k=10或k=-11(舍去),故a=2,k=10.(2)證明:由(1)得Sn=n(2+2n)則bn=Snn故bn+1-bn=(n+2)-(n+1)=1,且b1=2,即{bn}是首項為2,公差為1的等差數(shù)列,故Tn=n(能力提升1.(多選題)已知數(shù)列{an}為等差數(shù)列,其前n項和為Sn,若Sn=S13-n(n∈N*,且n<13),則下列結(jié)論正確的為()A.S13=0 B.a7=0C.{an}為遞增數(shù)列 D.a13=0答案:AB解析:對B,由題意,Sn=S13-n,令n=7,有S7=S6?S7-S6=0?a7=0,故B正確.對A,S13=13(a1+a13)2=13對C,當an=0時滿足Sn=S13-n=0,故{an}為遞增數(shù)列不一定正確.故C錯誤.對D,由A,B項,可設(shè)當an=7-n時滿足Sn=S13-n,但a13=-6.故D錯誤.故AB正確.2.已知等差數(shù)列{an}的前n項和為Sn,S4=40,Sn=210,Sn-4=130,則n等于()A.12 B.14 C.16 D.18答案:B解析:因為Sn-Sn-4=an+an-1+an-2+an-3=80,S4=a1+a2+a3+a4=40,所以4(a1+an)=120,a1+an=30.由Sn=n(a1+an3.在等差數(shù)列{an}中,若an=2n+3,其前n項和Sn=an2+bn+c(a,b,c為常數(shù)),則a-b+c=.

答案:-3解析:因為an=2n+3,所以a1=5,Sn=(5+2n+3)n2=n2+4n,與Sn=an2+bn+c比較,得a=1,b=4,4.已知等差數(shù)列{an}的前n項和為Sn,若OB=a1OA+a200OC,且A,B,C三點共線(該直線不過原點O),則S200=.

答案:100解析:因為A,B,C三點共線(該直線不過原點O),所以a1+a200=1,所以S200=200(a15.已知等差數(shù)列{an},{bn}的前n項和分別為Sn,Tn,且SnTn=7n+45n-答案:5解析:由等差數(shù)列的性質(zhì),知anbn=S2n-1T2n-1=7(2n-1)+45(2n6.已知數(shù)列{an}的所有項均為正數(shù),其前n項和為Sn,且Sn=14an2+(1)證明:{an}是等差數(shù)列;(2)求數(shù)列{an}的通項公式.(1)證明:當n=1時,a1=S1=14a12+解得a1=3或a1=-1(舍去).當n≥2時,an=Sn-Sn-1=14(an2+2an-3)-14(an-12+2an-1-3),得4an即(an+an-1)(an-an-1-2)=0.因為an+an-1>0,所以an-an-1=2(n≥2).故{an}是以3為首項,2為公差的等差數(shù)列.(2)解:由(1)知an=3+2(n-1)=2n+1.7.已知公差大于零的等差數(shù)列{an}的前n項和為Sn,且滿足a3a4=117,a2+a5=22.(1)求數(shù)列{an}的通項公式;(2)若{bn}是等差數(shù)列,且bn=Snn+c解:(1)設(shè)等差數(shù)列{an}的公差為d,則d>0.∵a3+a4=a2+a5=22,且a3a4=117,∴a3,a4是方程x2-22x+117=0的兩個根