4.2.2等差數(shù)列的前n項(xiàng)和公式第1課時(shí)等差數(shù)列的前n項(xiàng)和公式一．選擇題1.在等差數(shù)列{an}中,若a2=1,a4=5,則數(shù)列{an}的前5項(xiàng)和S5等于()A.7 B.15 C.20 D.252.已知等差數(shù)列{an}的前n項(xiàng)和Sn=n2+5n,則公差d等于()A.1 B.2 C.5 D.103.已知{an}是等差數(shù)列,a1=10,前10項(xiàng)和S10=70,則其公差d等于()A.-23 B.-13 C.13 4.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(xiàng)和,a2,a1021是函數(shù)f(x)=x2-6x+5的兩個(gè)零點(diǎn),則S1022的值為()A.3066 B.12 C.1020 D.30605.已知{an}是公差d=1的等差數(shù)列,Sn為數(shù)列{an}的前n項(xiàng)和,若S8=4S4,則a10等于()A.172 B.192 C.10 D6.(2023·全國甲高考)記Sn為等差數(shù)列{an}的前n項(xiàng)和.若a2+a6=10,a4a8=45,則S5=()A.25 B.22 C.20 D.157.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(xiàng)和為()A.160 B.180 C.200 D.2208.(多選題)已知等差數(shù)列10,7,4,…,則()A.該數(shù)列的通項(xiàng)公式為an=-3n+13B.-25是該數(shù)列的第13項(xiàng)C.該數(shù)列的前5項(xiàng)和最大D.設(shè)該數(shù)列為{an},則|a1|+|a2|+|a3|+…+|a8|=489.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.18二．填空題10.已知{an}是等差數(shù)列,Sn是它的前n項(xiàng)和.若S4=20,a4=8,則S8=.

11.在等差數(shù)列{an}中,若an=2n+3,其前n項(xiàng)和Sn=an2+bn+c(a,b,c為常數(shù)),則a-b+c=.

12.已知等差數(shù)列{an}的前n項(xiàng)和為Sn,若OB=a1OA+a200OC,且A,B,C三點(diǎn)共線(該直線不過原點(diǎn)O),則S200=.

13.已知等差數(shù)列{an},{bn}的前n項(xiàng)和分別為Sn,Tn,且SnTn=7n+45n三．解答題14.已知數(shù)列{an}滿足a1=50,an+1=an+2n.(1)求{an}的通項(xiàng)公式;(2)已知數(shù)列{bn}的前n項(xiàng)和為an,若bm=50,求正整數(shù)m的值.15.已知等差數(shù)列{an}的前三項(xiàng)依次為a,4,3a,前n項(xiàng)和為Sn,且Sk=110.(1)求a及k的值;(2)設(shè)數(shù)列{bn}的通項(xiàng)公式為bn=Snn,求證:數(shù)列{bn}是等差數(shù)列,并求其前n項(xiàng)和T16.已知數(shù)列{an}的所有項(xiàng)均為正數(shù),其前n項(xiàng)和為Sn,且Sn=14an2+(1)求證:{an}是等差數(shù)列;(2)求數(shù)列{an}的通項(xiàng)公式.17.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a3a4=117,a2+a5=22.(1)求數(shù)列{an}的通項(xiàng)公式;(2)若{bn}是等差數(shù)列,且bn=Snn+4.2.2等差數(shù)列的前n項(xiàng)和公式第1課時(shí)等差數(shù)列的前n項(xiàng)和公式一．選擇題1.B設(shè)等差數(shù)列{an}的公差為d,則有a故S5=5a1+5×42d=152.B∵a1=S1=6,a1+a2=S2=14,∴a2=8,∴d=a2-a1=2.3.A因?yàn)镾10=10a1+10×92d=70,a1=所以d=-234.A∵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,a1021是函數(shù)f(x)=x2-6x+5的兩個(gè)零點(diǎn),即a2,a1021是方程x2-6x+5=0的兩個(gè)根,∴a2+a1021=6,∴S1022=(a1+a15.B由題意,可知S8=8a1+8×72d=8a1+28,S4=4a1+4×32d=4a1+6.因?yàn)镾8=4S4,即8a1+28=4(4a1+6),解得a1=12,所以a10=a1+96.C設(shè)等差數(shù)列{an}的首項(xiàng)為a1,公差為d,由已知得2a1+6d=10,(a1+3d)(a7.B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.AD依題意知,首項(xiàng)a1=10,公差d=-3,所以該數(shù)列的通項(xiàng)公式為an=a1+(n-1)d=10-3(n-1)=-3n+13,故A正確;由an=-3n+13=-25,得n=383≠13,故B不正確由an=-3n+13≥0,得n≤4,由an=-3n+13<0,得n≥5,所以該數(shù)列的前4項(xiàng)和最大,故C不正確;數(shù)列{an}的前n項(xiàng)和Sn=10n+n(n-1)2×|a1|+|a2|+|a3|+…+|a8|=a1+a2+a3+a4-(a5+a6+a7+a8)=2S4-S8=2×-3×42+23×42?-3×89.B因?yàn)镾n-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+an二．填空題10.72設(shè)等差數(shù)列{an}的公差為d,則由S4=4a1+故S8=8×2+8×72×2=7211.-3因?yàn)閍n=2n+3,所以a1=5,Sn=(5+2n+3)n2=n2+4n,與Sn=an2+bn+c比較,得a=1,b=4,12.100因?yàn)锳,B,C三點(diǎn)共線(該直線不過原點(diǎn)O),所以a1+a200=1,所以S200=200(a113.5由等差數(shù)列的性質(zhì),知anbn=S2n-1T2n-1=7(2n-1)+45(2n三．解答題14.解:(1)當(dāng)n≥2時(shí),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(xiàng)公式為an=n2-n+50.(2)b1=a1=50,當(dāng)n≥2時(shí),bn=an-an-1=n2-n+50-[(n-1)2-(n-1)+50]=2n-2,即bn=50當(dāng)m≥2時(shí),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26.15.(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(xiàng)為2,公差為1的等差數(shù)列,故Tn=n(16.(1)證明:當(dāng)n=1時(shí),a1=S1=14a12+解得a1=3或a1=-1(舍去).當(dāng)n≥2時(shí),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?yàn)閍n+an-1>0,所以an-an-1=2(n≥2).故{an}是以3為首項(xiàng),2為公差的等差數(shù)列.(2)解:由(1)知an=3+2(n-1)=2n+1.17.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=