20xx屆高三數(shù)學(xué)理一輪復(fù)習(xí)考點(diǎn)規(guī)范練：第六章數(shù)列32word版含解析(參考版)
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【正文】 2n+16.即Sn=(2n3)2n+1+6.。2n+1=2+2(2n1)2n+(2n1)2n,∴Sn=2+322+…+(2n1)a5,即(a1+d)2=a122n2=nbn=(1)n1(1)n+1由等比數(shù)列{an}不是遞減數(shù)列,可得q=,故an==(1)n1(2)由bn=(1)n+1(an+1an).由于an0,可得an+1an=2.又+2a1=4a1+3,解得a1=1(舍去),a1=3.所以{an}是首項(xiàng)為3,公差為2的等差數(shù)列,故{an}的通項(xiàng)公式為an=2n+1.(2)由an=2n+1可知bn==設(shè)數(shù)列{bn}的前n項(xiàng)和為Tn,則Tn=b1+b2+…+bn= (1)因?yàn)?2Sn+n+4,所以=2Sn1+n1+4(n≥2).兩式相減,得=2an+1,所以+2an+1=(an+1)2.因?yàn)閧an}是各項(xiàng)均為正數(shù)的數(shù)列,所以an+1an=1.又=(a21)a7,所以(a2+1)2=(a21)(a2+5),解得a2=3,a1=2,所以{an}是以2為首項(xiàng),1為公差的等差數(shù)列,所以an=n+1.由題意知b1=2,b2=4,b3=8,故bn=2n.(2)由(1)得=(1)nlog22n=(1)nn,故Tn=c1+c2+…+=[1+23+…+(1)nn]設(shè)Fn=1+23+…+(1)nn.則當(dāng)n為偶數(shù)時(shí),Fn=(1+2)+(3+4)+…+[(n1)+n]=。2n1=2(n+4)21+21+22+23+…+2n(n+2)21+22+23+…+2n(n+2)22+423+…+(n+2)21+4(n+1)2=(1)n[n2(n+1)2]=(1)n+1n2,∴an=f(n)+f(n+1)=(1)n(2)若數(shù)列{bn}滿足an+log2n=log2bn,求數(shù)列{bn}的前n項(xiàng)和Tn.?導(dǎo)學(xué)號(hào)37270467?高考預(yù)測(cè){an}中,公差d≠0,a10=19,且a1,
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