第四章数列4.2等差数列4.2.2等差数列的前n项和公式第1课时等差数列的前n项和课后篇巩固提升基础达标练1.已知Sn为等差数列{an}的前n项和,a2+a5=4,S7=21,则a7的值为()A.6B.7C.8D.9解析设{an}的公差为d,则{a1+d+a1+4d=4,7a1+7×62d=21,解得{a1=-3,d=2,所以a7=a1+6d=-3+6×2=9,故选D.答案D2.(多选)(2020山东高三月考)记Sn为等差数列{an}的前n项和.若a4+a5=24,S6=48,则下列正确的是()A.a1=-2B.a1=2C.d=4D.d=-4解析因为{a4+a5=2a1+7d=24,S6=6a1+15d=48,所以{a1=-2,d=4.故选AC.答案AC3.已知数列{an}的前n项和Sn=n2,则an等于()A.nB.n2C.2n+1D.2n-1解析当n=1时,a1=S1=1;当n≥2时,an=Sn-Sn-1=n2-(n-1)2=2n-1,且a1=1适合上式,故an=2n-1(n∈N*).答案D4.在我国古代著名的数学专著《九章算术》里有一段叙述:今有良马与驽马发长安至齐,齐去安一千一百二十五里,良马初日行一百零三里,日增十三里;驽马初日行九十七里,日减半里,良马先至齐,复还迎驽马,二马相逢,问:相逢时良马比驽马多行()A.1125里B.920里C.820里D.540里解析设良马每天所行路程为{an},则{an}是以103为首项,以13为公差的等差数列,其前n项和为An,驽马每天所行路程为{bn},则{bn}是以97为首项,以-12为公差的等差数列,其前n项和为Bn,设共用n天二马相逢,则An+Bn=2×1125,所以103n+n(n-1)2×13+97n+n(n-1)2(-12)=2250,化简得n2+31n-360=0,解得n=9.A9=103×9+9×82×13=1395,B9=2250-1395=855,A9-B9=1395-855=540.答案D5.已知数列{an}的通项公式为an=2n+1,令bn=1n(a1+a2+…+an),则数列{bn}的前10项和T10=()A.70B.75C.80D.85解析 an=2n+1,∴数列{an}是等差数列,首项a1=3,其前n项和Sn=n(a1+an)2=n(3+2n+1)2=n2+2n,∴bn=1nSn=n+2,∴数列{bn}也是等差数列,首项b1=3,公差为1.∴其前10项和T10=10×3+10×92×1=75,故选B.答案B6.已知等差数列{an}中,a10=13,S9=27,则公差d=,a100=.解析S9=9a5=27⇒a5=3,d=a10-a55=13-35=2,∴a100=a10+90d=13+90×2=193.答案21937.(2019全国Ⅲ,理14)记Sn为等差数列{an}的前n项和.若a1≠0,a2=3a1,则S10S5=.解析设等差数列{an}的公差为d. a1≠0,a2=3a1,∴a1+d=3a1,即d=2a1.∴S10S5=10a1+10×92d5a1+5×42d=100a125a1=4.答案48.已知数列{an}的前n项和为Sn=n·2n-1,则a3+a4+a5=.解析a3+a4+a5=S5-S2=(5×25-1)-(2×22-1)=152.答案1529.设数列{an}的前n项和为Sn,点(n,Snn)(n∈N*)均在函数y=3x-2的图象上,求数列{an}的通项公式.解依题意,得Snn=3n-2,即Sn=3n2-2n.当n≥2时,an=Sn-Sn-1=(3n2-2n)-[3(n-1)2-2(n-1)]=6n-5.因为a1=S1=1,满足an=6n-5,所以an=6n-5(n∈N*).10.已知数列{an}满足a1=1,a2=2,an+2=2an+1-an+2.(1)设bn=an+1-an,证明{bn}是等差数列;(2)求{an}的通项公式.解(1) an+2=2an+1-an+2,∴an+2-an+1=an+1-an+2,即bn+1=bn+2.又b1=a2-a1=2-1=1,∴数列{bn}是以1为首项,2为公差的等差数列.(2)由(1)可知,an+1-an=1+2(n-1)=2n-1,∴an-an-1=2(n-1)-1,an-1-an-2=2(n-2)-1,……a2-a1=2×1-1,累加,得an-a1=2×n(n-1)2-(n-1)=n2-2n+1,∴an=a1+n2-2n+1=n2-2n+2,∴数列{an}的通项公式为an=n2-2n+2.能力提升练1.在等差数列{an}中,2a4+a7=3,则数列{an}的前9项和S9等于()A.3B.6C.9D.12解析设等差数列{an}的公差为d,因为2a4+a7=3,所以2(a1+3d)+a1+6d=3,整理,得a1+4d=1,即a5=1,所以S9=9(a1+a9)2=9a5=9.答案C2.若公差不为0的等差数列{an}的前21项的和等于前8项的和,且a8+ak=0,则正整数k的值为()A.20B.21C.22D.23解析设等差数列{an}的前n项和为Sn,由题意,得S21=S8,即a9+a10+…+a21=0.根据等差数列的性质,得13a15=0,即a15=0.故a8+a22=2a15=0,即k=22.故选C.答案C3.已知等差数列{an},a2=6,a5=15,若bn=a2n,则数列{bn}的前5项和等于()A.30B.45C.90D.186解析由等差数列{an}易得公差d1=3.又bn=a2n,所以{bn}也是等差数列,公差d2=6.故S5=b1+b2+b3+b4+b5=a2+a4+a6+a8+a10=5×6+5×42×6=90.答案C4.(2020河北正定中学高一月考)设等差数列{an}的前n项和是Sn,已知S14>0,S15<0,下列选项正确的是()A.a1>0,d<0B.a7+a8>0C.S6与S7均为Sn的最大值D.a8<0解析根据题意,有S14=14×(a1+a14)2=7(a1+a14)=7(a7+a8)>0,即a7+a8>0,S15=15×(a1+a15)2=15a8<0,即a8<0,则a7...
