20xx高考數(shù)學(xué)文人教a版一輪復(fù)習(xí)學(xué)案：64-數(shù)列求和-【含解析】-資料下載頁
2025-04-03 02:51本頁面
【正文】 =2n+1.由已知可得an+1(2n+3)=3[an(2n+1)],an(2n+1)=3[an1(2n1)],……a25=3(a13).因?yàn)閍1=3,所以an=2n+1.(2)由(1)得2nan=(2n+1)2n,所以8Sn=32+522+723+…+(2n+1)2n.①從而2Sn=322+523+724+…+(2n+1)2n+1.②①②得Sn=32+222+223+…+22n(2n+1)2n+1.所以Sn=(2n1)2n+1+2.對點(diǎn)訓(xùn)練3解(1)設(shè){an}的公比為q,由題設(shè)得2a1=a2+a3,即2a1=a1q+a1q2.所以q2+q2=0,解得q=1(舍去),q={an}的公比為2.(2)記Sn為{nan}的前n項(xiàng)和.由(1)及題設(shè)可得,an=(2)n1.所以Sn=1+2(2)+…+n(2)n1,2Sn=2+2(2)2+…+(n1)(2)n1+n(2)n.可得3Sn=1+(2)+(2)2+…+(2)n1n(2)n=1(2)n3n(2)n.所以Sn=19(3n+1)(2)n9.例4(1)解由Sn=12an+1+n+1(n∈N*),得Sn1=12an+n(n≥2,n∈N*),兩式相減,并化簡得an+1=3an2,即an+11=3(an1),又a11=21=3≠0,∴數(shù)列{an1}是以3為首項(xiàng),3為公比的等比數(shù)列,∴an1=(3)3n1==3n+1.(2)證明由bn=log3(an+1)=log33n=n,得1bnbn+2=1n(n+2)=121n1n+2,∴Tn=12113+1214+1315+…+1n11n+1+1n1n+2=121+121n+11n+2=342n+32(n+1)(n+2)34.對點(diǎn)訓(xùn)練4解(1)令n=1,得a1b1=3+(23)2=1,所以b1=1,令n=2,得a1b1+a2b2=7,所以a2b2==3,所以a2=2,設(shè)數(shù)列{an}的公比為q,則q=a2a1=2.所以an=2n1.(2)當(dāng)n≥2時(shí),a1b1+a2b2+…+an1bn1=3+(2n5)2n1,①又a1b1+a2b2+a3b3+…+anbn=3+(2n3)2n,②②①,得anbn=3+(2n3)2n[3+(2n5)2n1]=(2n1)2n1,得bn=2n1,當(dāng)n=1時(shí)也成立,所以bn=2n1,1bnbn+1=1(2n1)(2n+1)=1212n112n+1.所以Tn=12113+121315+…+1212n112n+1=12113+1315+…+12n112n+1=12112n+1=n2n+1.8
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