20xx高考數(shù)學文人教a版一輪復習學案：64-數(shù)列求和-【含解析】(參考版)
2025-04-03 02:51本頁面
【正文】 (2n1).當n為偶數(shù)時,Tn=13+57+…+(2n3)(2n1)=n.當n為奇數(shù)時,Tn=Tn1+bn=(n1)+(1)n1(2n1)=(n1)+(2n1)=,Tn=(1)n+1n.例3解(1)a2=5,a3=7.猜想an=2n+1.由已知可得an+1(2n+3)=3[an(2n+1)],an(2n+1)=3[an1(2n1)],……a25=3(a13).因為a1=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.對點訓練3解(1)設{an}的公比為q,由題設得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項和.由(1)及題設可得,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,3為公比的等比數(shù)列,∴an1=(3)log22n=(1)nlog223(32n1,即有an=3n]13414114+34134=103.對點訓練1解(1)設等差數(shù)列{an}的公差為d,d0,由已知,得an=1+(n1)d,a22=a1a5,即(1+d)2=1+4d,解得d=2或d=0(舍去),所以an=2n1,n∈N*。學案突破例1證明(1)由題意,得2an+1=an+12bn,2bn+1=12an+bn,兩式相加,得an+1+bn+1=34(an+bn),∵a1=1,b1=12,∴a1+b1=32,∴{an+bn}是首項為32,公比為34的等比數(shù)列.兩式相減,得an+1bn+1=14(anbn),∵a1b1=12,∴{anbn}是首項為12,公比為14的等比數(shù)列.(2)由(1)可得an+bn=3234n1,①anbn=1214n1,②兩式相加,得an=14n+34n,∴Sn=14[(1
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