新教材適用2023-2024學(xué)年高中數(shù)學(xué)第一章數(shù)列2等差數(shù)列2.2等差數(shù)列的前n項(xiàng)和第2課時(shí)等差數(shù)列前n項(xiàng)和的性質(zhì)課后訓(xùn)練北師大版選擇性必修第二冊(cè)

第2課時(shí)等差數(shù)列前n項(xiàng)和的性質(zhì)課后訓(xùn)練鞏固提升A組1.已知等差數(shù)列{an}的前n項(xiàng)和為Sn,若S2=4,S4=20,則該數(shù)列的公差d=().A.7 B.6 C.3 D.2解析:∵S4-S2=a3+a4=20-4=16,S2=a1+a2,∴(a3+a4)-(a1+a2)=16-4=4d,∴d=3.答案:C2.已知數(shù)列{an}是等差數(shù)列,且a2=-8,a15=5,Sn是數(shù)列{an}的前n項(xiàng)和,則().A.S10=S11 B.S10>S11 C.S9=S10 D.S9<S10解析:∵d=a15-a215-2=令an=0,得n=10,即a10=0.∴S9=S10.由d=1,得a11=1,故S11>S10.答案:C3.已知等差數(shù)列{an}的前n項(xiàng)和為Sn,若S3=9,S6=36,則a7+a8+a9=().A.63 B.45 C.36 D.27解析:∵數(shù)列{an}是等差數(shù)列,Sn是其前n項(xiàng)和,∴S3,S6-S3,S9-S6成等差數(shù)列.∵S3=9,S6-S3=27,∴S9-S6=45.∵a7+a8+a9=S9-S6,∴a7+a8+a9=45.答案:B4.已知等差數(shù)列{an}的前n項(xiàng)和為Sn,若S3=-6,S18-S15=18,則S18=.

解析:∵S3=a1+a2+a3=-6,S18-S15=a18+a17+a16=18,∴(a1+a18)+(a2+a17)+(a3+a16)=12.即a1+a18=4,∴S18=18(a1答案:365.已知等差數(shù)列{an}的前n項(xiàng)和為Sn,S10=0,S15=25,則Sn的最小值為.

解析由Sn=na1+n(n-解得a1=-3,d=23,∴Sn=-3n+n(n-1)2∴當(dāng)n=5時(shí),Sn有最小值-253答案:-256.已知等差數(shù)列{an}的前n項(xiàng)和為Sn,且a4-a2=8,a3+a5=26,記Tn=Snn2,若存在正整數(shù)M,使得對(duì)一切正整數(shù)n,Tn≤M都成立,則M的最小值是解析:∵a4-a2=8,∴a5-a3=8.又a3+a5=26,∴a3=9,a5=17.∴an=4n-3,Sn=n(1+4n-3∴Snn2=2n-1n=2-1n,∴M≥2,∴M的最小值為2.答案:27.在等差數(shù)列{an}中,a10=18,前5項(xiàng)的和S5=-15.(1)求數(shù)列{an}的通項(xiàng)公式;(2)求數(shù)列{an}的前n項(xiàng)和的最小值,并指出何時(shí)取得最小值.解:(1)設(shè){an}的首項(xiàng)、公差分別為a1,d.則a1+9∴an=3n-12.(2)Sn=n(a1+an)2=12(3n2-21n)=32n-722-1478,當(dāng)n=3或8.已知數(shù)列{an}為等差數(shù)列,其前n項(xiàng)和為Sn,且S4=-62,S6=-75,求:(1){an}的通項(xiàng)公式及前n項(xiàng)和Sn;(2)求|a1|+|a2|+…+|a14|的值.解:(1)設(shè)等差數(shù)列{an}的公差為d,則4a1∴an=3n-23,Sn=32n2-432(2)由an=3n-23,得當(dāng)n≤7時(shí),an<0,當(dāng)n>7時(shí),an>0,∴|a1|+|a2|+…+|a14|=-(a1+a2+…+a7)+(a8+a9+…+a14)=-S7+S14-S7=S14-2S7=147.B組1.已知數(shù)列{an}的前n項(xiàng)和為Sn,且an=-2n+1,則數(shù)列Snn的前11項(xiàng)和為(A.-45 B.-50 C.-55 D.-66解析:∵Sn=n(-1-2n+1)2=-n2,∴S∴Snn的前11項(xiàng)和為-答案:D2.已知{an}為等差數(shù)列,若a11a10<-1,且它的前n項(xiàng)和Sn有最大值,則當(dāng)Sn取得最小正值時(shí),n=(A.11 B.19 C.17 D.21解析:∵Sn有最大值,∴{an}的公差d<0.∵a11a10<-1,∴a11<0,a10>0,∴a10+a11∴S20=20(a1+a20)2=10(a10+a11)<0,S19=∴當(dāng)Sn取最小正值時(shí),n=19.答案:B3.若數(shù)列{an}的前n項(xiàng)和Sn=3n-2n2(n∈N+),則當(dāng)n≥2時(shí),下列不等式成立的是().A.Sn>na1>nan B.Sn>nan>na1C.na1>Sn>nan D.nan>Sn>na1解析:由an=S1,n=1,所以an=5-4n,所以na1=n,nan=5n-4n2.因?yàn)閚a1-Sn=n-(3n-2n2)=2n2-2n=2n(n-1)>0,Sn-nan=3n-2n2-(5n-4n2)=2n2-2n>0,所以na1>Sn>nan.答案:C4.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=().A.12 B.14 C.16 D.18解析:Sn-Sn-4=an+an-1+an-2+an-3=80,S4=a1+a2+a3+a4=40,所以4(a1+an)=120,即a1+an=30,由Sn=n(a1+an答案:B5.設(shè)等差數(shù)列{an}的前n項(xiàng)和為Sn,若a2=-3,S5=-10,則a5=;Sn的最小值為.

解析:設(shè)首項(xiàng)為a1,公差為d,由題意得a2=所以a5=a1+4d=0.an=a1+(n-1)d=n-5.令an≤0,則n≤5,即數(shù)列{an}中前4項(xiàng)為負(fù),a5=0,第6項(xiàng)及以后的項(xiàng)為正.故Sn的最小值為S4=S5=-10.答案:0-106.已知首項(xiàng)為正數(shù)的等差數(shù)列的前n項(xiàng)和為Sn,且S3=S8,當(dāng)n=時(shí),Sn取到最大值.

解析:∵S3=S8,∴S8-S3=a4+a5+a6+a7+a8=5a6=0,∴a6=0.∵a1>0,∴a1>a2>a3>a4>a5>a6=0,a7<0.故當(dāng)n=5或n=6時(shí),Sn最大.答案:5或67.項(xiàng)數(shù)為奇數(shù)的等差數(shù)列,奇數(shù)項(xiàng)之和為44,偶數(shù)項(xiàng)之和為33,求這個(gè)數(shù)列的中間項(xiàng)及項(xiàng)數(shù).解:設(shè)等差數(shù)列共(2n+1)項(xiàng),則奇數(shù)項(xiàng)有(n+1)項(xiàng),偶數(shù)項(xiàng)有n項(xiàng),中間項(xiàng)是第n+1項(xiàng),記為an+1,設(shè)公差為d,則S∴S奇-S偶=a1+nd=an+1=11,即中間項(xiàng)an+1=11.又S2n+1=S奇+S偶=77,∴(2n∴(2n+1)×11=77,∴2n+1=7,即數(shù)列的中間項(xiàng)為11,這個(gè)數(shù)列共7項(xiàng).8.已知數(shù)列{an}的前n項(xiàng)和為Sn,數(shù)列{an}為等差數(shù)列,a1=12,d=-2.(1)求Sn,并畫(huà)出{Sn}(1≤n≤13)的圖象;(2)分別求使Sn單調(diào)遞增、單調(diào)遞減的n的取值范圍,并求Sn的最大(或最小)的項(xiàng);(3){Sn}有多少項(xiàng)大于零?解:(1)Sn=na1