2020版高考數(shù)學(xué)一輪復(fù)習(xí) 課時規(guī)范練29 等差數(shù)列及其前n項(xiàng)和 理 北師大版.doc

課時規(guī)范練29等差數(shù)列及其前n項(xiàng)和基礎(chǔ)鞏固組1.由a1=1,d=3確定的等差數(shù)列an,當(dāng)an=298時,序號n等于()A.99B.100C.96D.1012.(2018湖南長郡中學(xué)仿真,6)已知等差數(shù)列an滿足an+1+an=4n,則a1=()A.-1B.1C.2D.33.(2018河南商丘二模,3)已知等差數(shù)列an的公差為d,且a8+a9+a10=24,則a1d的最大值為()A.B.C.2D.44.在等差數(shù)列an中,a3+a6=11,a5+a8=39,則公差d為()A.-14B.-7C.7D.145.已知等差數(shù)列an的前n項(xiàng)和為Sn,且3a3=a6+4,若S5<10,則a2的取值范圍是()A.(-,2)B.(-,0)C.(1,+)D.(0,2)6.已知Sn是等差數(shù)列an的前n項(xiàng)和,則2(a1+a3+a5)+3(a8+a10)=36,則S11=()A.66B.55C.44D.337.(2018湖南衡陽一模,15)已知數(shù)列an前n項(xiàng)和為Sn,若Sn=2an-2n,則Sn=.8.設(shè)數(shù)列anbn都是等差數(shù)列,若a1+b1=7,a3+b3=21,則a5+b5=.9.若數(shù)列an的前n項(xiàng)和為Sn,且滿足an+2SnSn-1=0(n2),a1=.(1)求證:1Sn成等差數(shù)列;(2)求數(shù)列an的通項(xiàng)公式.10.設(shè)數(shù)列an的前n項(xiàng)和為Sn,且Sn=2n-1.數(shù)列bn滿足b1=2,bn+1-2bn=8an.(1)求數(shù)列an的通項(xiàng)公式;(2)證明:數(shù)列bn2n為等差數(shù)列,并求bn的通項(xiàng)公式.綜合提升組11.(2018河北衡水中學(xué)考前押題二,10)已知數(shù)列a1=1,a2=2,且an+2-an=2-2(-1)n,nN+,則S2 017的值為()A.2 0161 010-1B.1 0092 017C.2 0171 010-1D.1 0092 01612.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)值最大時,n的值為()A.6B.7C.8D.913.已知等差數(shù)列an的前n項(xiàng)和為Sn,若Sm-1=-4,Sm=0,Sm+2=14(m2,且mN+),則m的值為.14.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)求通項(xiàng)公式an;(2)求Sn的最小值;(3)若數(shù)列bn是等差數(shù)列,且bn=Snn+c,求非零常數(shù)c.創(chuàng)新應(yīng)用組15.(2018湖南長郡中學(xué)仿真,15)若數(shù)列an是正項(xiàng)數(shù)列,且a1+a2+an=n2+3n,則a12+a23+ann+1=.16.等差數(shù)列an的前n項(xiàng)和為Sn,已知S10=0,S15=25,則nSn的最小值為多少?參考答案課時規(guī)范練29等差數(shù)列及其前n項(xiàng)和1.B根據(jù)等差數(shù)列通項(xiàng)公式an=a1+(n-1)d,有298=1+(n-1)3,解得n=100,故選B.2.B由題意,當(dāng)n分別取1,2時,a1+a2=4,a3+a2=8,解得公差d=2,故a1=1.故選B.3.Ca8+a9+a10=24,a9=8,即a1+8d=8,a1=8-8d,a1d=(8-8d)d=-8d-122+22,當(dāng)d=12時,a1d的最大值為2,故選C.4.Ca3+a6=11,a5+a8=39,則4d=28,解得d=7.故選C.5.A設(shè)公差為d,由3a3=a6+4得3(a2+d)=a2+4d+4,即d=2a2-4,由S5<10得,5(a1+a5)2=5(a2+a4)2=5(2a2+2d)2=5(3a2-4)<10,解得a2<2,故選A.6.D由等差數(shù)列的性質(zhì)可得2(a1+a3+a5)+3(a8+a10)=6a3+6a9=36,即a1+a11=6.則S11=11(a1+a11)2=113=33.故選D.7.n2nSn=2an-2n=2(Sn-Sn-1)-2n,整理得Sn-2Sn-1=2n,等式兩邊同時除以2n,則Sn2n-Sn-12n-1=1.又S1=2a1-2=a1,可得a1=S1=2,數(shù)列Sn2n是以1為首項(xiàng),公差為1的等差數(shù)列,Sn2n=n,Sn=n2n.8.35數(shù)列an,bn都是等差數(shù)列,設(shè)數(shù)列an的公差為d1,數(shù)列bn的公差為d2,a3+b3=a1+b1+2(d1+d2)=21,而a1+b1=7,可得2(d1+d2)=21-7=14.a5+b5=a3+b3+2(d1+d2)=21+14=35.9.(1)證明 當(dāng)n2時,由an+2SnSn-1=0,得Sn-Sn-1=-2SnSn-1,1Sn-1Sn-1=2.又1S1=1a1=2,故1Sn是首項(xiàng)為2,公差為2的等差數(shù)列.(2)解 由(1)可得1Sn=2n,Sn=12n.當(dāng)n2時,an=Sn-Sn-1=12n-12(n-1)=n-1-n2n(n-1)=-12n(n-1).當(dāng)n=1時,a1=12不適合上式.故an=12,n=1,-12n(n-1),n2.10.(1)解 當(dāng)n=1時,a1=S1=21-1=1;當(dāng)n2時,an=Sn-Sn-1=(2n-1)-(2n-1-1)=2n-1.a1=1適合通項(xiàng)公式an=2n-1,an=2n-1.(2)證明 bn+1-2bn=8an,bn+1-2bn=2n+2,即bn+12n+1-bn2n=2.又b121=1,bn2n是首項(xiàng)為1,公差為2的等差數(shù)列.bn2n=1+2(n-1)=2n-1.bn=(2n-1)2n.11.C由題意,當(dāng)n為奇數(shù)時,an+2-an=4,數(shù)列a2n-1是首項(xiàng)為1,公差為4的等差數(shù)列,當(dāng)n為偶數(shù)時,an+2-an=0,數(shù)列a2n-1是首項(xiàng)為2,公差為0的等差數(shù)列,S2 017=(a1+a3+a2 017)+(a2+a4+a2 016)=1 009+1 0091 0084+1 0082=2 0171 010-1,故選C.12.Ba1=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)數(shù)列an的前k項(xiàng)和數(shù)值最大,則有ak0,ak+10,kN+.22-3k0,22-3(k+1)0.193k223.kN+,k=7.滿足條件的n的值為7.13.5Sm-1=-4,Sm=0,Sm+2=14,am=Sm-Sm-1=4,am+1+am+2=Sm+2-Sm=14.設(shè)數(shù)列an的公差為d,則2am+3d=14,d=2.Sm=a1+am2m=0,a1=-am=-4.am=a1+(m-1)d=-4+2(m-1)=4,m=5.14.解 (1)數(shù)列an為等差數(shù)列,a3+a4=a2+a5=22.又a3a4=117,a3,a4是方程x2-22x+117=0的兩實(shí)根.又公差d>0,a3<a4,a3=9,a4=13,a1+2d=9,a1+3d=13,a1=1,d=4.通項(xiàng)公式an=4n-3.(2)由(1)知a1=1,d=4,Sn=na1+n(n-1)2d=2n2-n=2n-142-18.當(dāng)n=1時,Sn最小,最小值為S1=a1=1.(3)由(2)知Sn=2n2-n,bn=Snn+c=2n2-nn+c,b1=11+c,b2=62+c,b3=153+c.數(shù)列bn是等差數(shù)列,2b2=b1+b3,即62+c2=11+c+153+c,2c2+c=0,c=-12(c=0舍去),故c=-12.15.2n2+6n由a1+a2+an=n2+3n,則a1+a2+an-1=(n-1)2+3(n-1),兩式相減,可得an=2n+2,當(dāng)n=1時也成立,則an=(2n+2)2,有ann+1=(2n+2)2n+1=4n+4,ann+1為公差為4的等差數(shù)列,其前n項(xiàng)和為a12+a23+ann+1=na12+ann+12=n(8+4n+4)2=2n2+6n.16.解 設(shè)數(shù)列an的首項(xiàng)為a1,公差為d,則S10=10a1+1092d=10a1+45d=0,S15=15a1+15142d=15a1+105d=25.聯(lián)立,得a1=-3,d=23,Sn=-3n+n(n-1)223=13n2-103n.令f(n)=nSn,則f(n)=13n3-103n2,f(n)=n2-203n.令f(n)=0,得n=0或n=203.當(dāng)n>203時,f(n)>0,當(dāng)0<n<203時,f(n)<0,當(dāng)n=203時,f(n)取最小值,又nN+,f(6)=-48,f(7)=-49,當(dāng)n=7時,f(n)取最小值-49.