20xx高考數(shù)學(xué)文人教a版一輪復(fù)習(xí)學(xué)案：62-等差數(shù)列及其前n項(xiàng)和-【含解析】-資料下載頁(yè)
2025-04-03 03:18本頁(yè)面
【正文】 1a3=5=5d,即d=1,所以1an=1a3+(n3)d=5+(n3)1=n+2,故an=1n+2.(2)解由(1)可知bn=an2n+6=121(n+2)(n+3)=121n+21n+3,故Sn=121314+1415+…+1n+21n+3=12131n+3=n6n+18.例3(1)A (2)72 (1)a7b7=2a72b7=a1+a13b1+b13=a1+a13213b1+b13213=S13T13=3132213+1=3727.(2)由S3=S9,得a4+a5+a6+a7+a8+a9=0,所以a6+a7=0,又因?yàn)閍2+a5=240,所以a1+a6=240,因?yàn)榈炔顢?shù)列的通項(xiàng)公式an是關(guān)于n的單調(diào)函數(shù),所以數(shù)列{an}的前6項(xiàng)為正,所以當(dāng)n=6時(shí),Sn有最大值,且S6=3(a1+a6)=324=72.例4(1)210 (2)A (1)記數(shù)列{an}的前n項(xiàng)和為Sn,由等差數(shù)列前n項(xiàng)和的性質(zhì)知Sm,S2mSm,S3mS2m成等差數(shù)列,則2(S2mSm)=Sm+(S3mS2m).又Sm=30,S2m=100,S2mSm=10030=70,∴S3mS2m=2(S2mSm)Sm=110,∴S3m=110+100=210.(2)由于點(diǎn)(n,an)(n∈N*)在函數(shù)y=lnx的圖象上,則an=lnn,則ean=n,所以Sn=ea1+ea2+…+ean=1+2+…+n=n(n+1)2,所以Sn是關(guān)于n的增函數(shù),由于滿足Sn=ea1+ea2+…+ean≥m的n的最小值為5,則S4m≤S5,所以10m≤(10,15].故選A.對(duì)點(diǎn)訓(xùn)練3(1)C (2)A (3)D (1)等差數(shù)列{an}的前n項(xiàng)和為Sn,若S4=20,S5=30,進(jìn)而得到a5=10,根據(jù)等差數(shù)列的性質(zhì)得到S5=30=5a3,故a3=6,根據(jù)等差數(shù)列的性質(zhì)得到a5a3=2d,d=2,am=40=a5+2(m5),解得m=.(2)因?yàn)樵诘炔顢?shù)列{an}中,a2+a8=8,所以a2+a8=2a5=8,解得a5=4,(a3+a7)2a5=(2a5)2a5=644=60.(3)因?yàn)楣頳0,(S8S5)(S9S5)0,所以S9S8,所以S8S5S9,所以a6+a7+a80,a6+a7+a8+a90,即3a70,2(a7+a8)0,所以|a7||a8|.例5解(方法1)由S3=S11,得3a1+322d=11a1+11102d,解得d=213a1,從而Sn=d2n2+a1d2n=a113(n7)2+4913a1,因?yàn)閍10,所以a1130.故當(dāng)n=7時(shí),Sn最大.(方法2)由解法1可知,d=213a1.要使Sn最大,則有an≥0,an+1≤0,即a1+(n1)(213a1)≥0,a1+n(213a1)≤0,≤n≤,故當(dāng)n=7時(shí),Sn最大.(方法3)由S3=S11,可得2a1+13d=0,即(a1+6d)+(a1+7d)=0,故a7+a8=0,又由a10,S3=S11可知d0,所以a70,a80,所以當(dāng)n=7時(shí),Sn最大.對(duì)點(diǎn)訓(xùn)練4解(1)設(shè)等差數(shù)列{an}的公差為d,由a3=9,a10=5,得a1+2d=9,a1+9d=5,解得a1=13,d=2.∴數(shù)列{an}的通項(xiàng)公式為an=2n15.(2)由(1)得Sn=13n+n(n1)22=n214n,∵Sn=(n7)249,∴當(dāng)n=7時(shí),Sn取得最小值.9
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