1、考点突破练5数列求和方法及综合应用1.(2022陕西宝鸡三模)已知数列an中,a1=a2=1,且an+2=an+1+2an.记bn=an+1+an.(1)求证:数列bn是等比数列;(2)求数列bn+2n的前n项和.2.(2022黑龙江哈尔滨第九中学二模)已知数列an满足a1a2an=2-2an,nN*.(1)证明:数列11-an是等差数列,并求数列an的通项公式;(2)记bn=ann(n+1),求bn的前n项和Sn.3.(2022安徽合肥二模)从cn=an+2(an+1-1)(an+1-2),cn=log2an+2an+1两个条件中任选一个,补充在第(2)问中的横线上并作答.记Sn为数列an的
2、前n项和,已知a1=1,且Sn=an+1-3.(1)求数列an的通项公式;(2)已知数列cn满足,记Tn为数列cn的前n项和,证明:Tn0对一切正奇数n恒成立,求实数m的取值范围.考点突破练5数列求和方法及综合应用1.(1)证明 由题可知an0,bn0.bn+1bn=an+2+an+1an+1+an=an+1+2an+an+1an+1+an=2(an+1+an)an+1+an=2,且b1=a1+a2=2,bn是以2为首项,2为公比的等比数列.(2)解 由(1)知,bn=2n,则bn+2n=2n+2n.设bn+2n的前n项和为Sn,则Sn=2(1-2n)1-2+n(2+2n)2=2n+1+n2+
3、n-2.2.(1)证明 由题可知an0.当n=1时,a1=2-2a1,得a1=23.当n2时,有a1a2an=2-2an,a1a2an-1=2-2an-1,an=1-an1-an-1,整理得11-an-1=an1-an=11-an-1,即11-an-11-an-1=1,11-an为首项为11-a1=3,公差为1的等差数列,11-an=3+(n-1)=n+2,即an=n+1n+2.(2)解 由(1)可得bn=ann(n+1)=1n(n+2)=121n-1n+2,Sn=121-13+12-14+13-15+1n-1-1n+1+1n-1n+2=121+12-1n+1-1n+2=34-2n+32(n+
4、1)(n+2).3.(1)解 由题可知Sn=an+1-3.当n=1时,a1=a2-3,a2=4.当n2时,Sn-1=an-3,-得an=(an+1-3)-(an-3)=an+1-an,即an+1=2an.又a2a1=42,数列an从第2项起成等比数列,即当n2时,an=a22n-2=2n,an=1,n=1,2n,n2.(2)证明 若选择:由(1)可得cn=an+2(an+1-1)(an+1-2)=2n+2(2n+1-1)(2n+1-2)=22n(2n+1-1)(2n-1)=212n-1-12n+1-1,Tn=21-122-1+122-1-123-1+12n-1-12n+1-1=21-12n+1-12.若选择:由(1)可得cn=n+22n+1,则Tn=322+423+n+12n+n+22n+1,12Tn=323+424+n+12n+1+n+22n+2,-得12Tn=34+123+124+12n+1-n+22n+2=34+141-12n-1-n+22n+2,Tn=2-n+42n+10对一切正奇数n恒成立,
