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中档大题满分练3.数列(A组)中档大题集训练,练就慧眼和规范,筑牢高考满分根基!1.已知公差不为0的等差数列an的前n项和为Sn,a1=a(aR),a1,a2,a4成等比数列.(1)求数列an的通项公式.(2)记1Sn的前n项和为An,1a2n-1的前n项和为Bn,当n2时,判断An与Bn的大小.【解析】(1)设an的公差为d,则由a1,a2,a4成等比数列,得a1=a0且a22=a1a4,所以(a1+d)2=a1(a1+3d).因为d0,所以解得d=a1=a,所以an=na.(2)由(1)得Sn=n(n+1)a2,所以1Sn=2a1n-1n+1.所以An=2a1-12+12-13+1n-1n+1=2a1-1n+1,又因为a2n-1=2n-1a,1a2n-1=1a12n-1,所以Bn=1a1+12+122+12n-1=1a1-12n1-12=2a1-12n.当n2时,2n=Cn0+Cn1+Cnn1+n0,即1-1n+10时,AnBn;当aBn.2.已知数列an满足an+1=2an+2n+1,且a1=2.(1)证明:数列an2n是等差数列.(2)设数列cn=ann-log2ann,求数列cn的前n项和Sn.【解析】(1)方法一:an+12n+1-an2n=2an+2n+12n+1-an2n=2an2n+1+2n+12n+1-an2n=1,且a121=1.所以数列an2n是以1为首项,公差为1的等差数列.方法二:由已知,an+1=2an+2n+1两边除以2n+1得an+12n+1=2an2n+1+1,即an+12n+1-an2n=1,又a121=1.所以数列an2n是以1为首项,公差为1的等差数列.(2)由(1)得an2n=1+(n-1)1=n,故an=n2n.所以cn=2n-n.所以Sn=c1+c2+c3+cn=(21-1)+(22-2)+(23-3)+(2n-n)=(21+22+23+2n)-(1+2+3+n)=2(1-2n)1-2-(1+n)n2=2n+1-n(n+1)2-2.故数列cn的前n项和为Sn=2n+1-n(n+1)2-2.
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