数列an各项均为正数sn为其前n项和总有an,sn,an^2成等差数列
来源:学生作业帮助网 编辑:作业帮 时间:2024/05/03 08:43:39
数列各项均为正,Sn>0.2√Sn是a(n+2)与an的等比中项,则(2√Sn)²=(an+2)an4Sn=an²+2ann=1时,4a1=4S1=a1²+2a1a1
2Sn=an+an^22Sn-1=an-1+an-1^2两式相减:2an=an^2+an-(an-1+an-1^2)an^2-an-(an-1+an-1^2)=0(an-(an-1+1))(an+an
4Sn=(an+1)^24Sn-1=(an-1+1)^2n-1为下标则4an=4Sn-4Sn-1=(an+1)^2-(an-1+1)^2化简得(an-1)^2=(an-1+1)^2则an-1=正负(a
an+Sn=4a(n-1)+S(n-1)=4相减:an/a(n-1)=1/2等比数列n=1时a1+a1=4a1=2an=2^(2-n)bn=1/n²数学归纳法n=2时T2=5/4
n=1时,2S1=2a1=a1²+a1a1²-a1=0a1(a1-1)=0a1=0(各项均为正数,舍去)或a1=1n≥2时,2Sn=an²+an2Sn-1=a(n-1)&
n=1,a1=2,n>1,an-1+Sn-1=4(1)an-Sn=4(2)(2)-(1)an/an-1=1/2,an=2x(1/2)^(n-1)=0.5^(n-2)n为正整数(2)补充问题你在检查一下
(1)(an+2)/2=根号下2Sn所以8Sn=(an+2)^2n=1,S1=a1.8a1=(a1+2)^2,得a1=2n=2,8S2=(a2+2)^2,8(a1+a2)=(a2+2)^2,得a2=6
1.n=1时,2a1=2S1=a1²+1-4a1²-2a1-3=0(a1+1)(a1-3)=0a1=-1(数列各项均为正,舍去)或a1=3n≥2时,2an=2Sn-2S(n-1)=
第一部分:补充:an=2n稍后上传第二部分再问:嗯嗯再答:稍微有点复杂,但是肯定没错。可能有更好的拆法,等老师评讲的时候仔细听一下。5/3=1500/900看我算的这么辛苦,采纳一下吧!
1)6Sn=An^2+3An+2因为S1=A1所以6A1=A1^2+3A1+2A1^2-3A1+2=0(A1-1)(A1-2)=0因为A1=S1>1所以A1=2因为An=Sn-S(n-1)注S(n-1
当n=1时,S1=a1=1/2(a1^2+a1),解得a1=1当n>1时,an=Sn-S(n-1)=1/2(an^2+an)-1/2[a(n-1)^2+a(n-1)],整理得[an+a(n-1)][a
4Sn=2an+an^24S(n-1)=2a(n-1)+a(n-1)^2相减得4an=2an-2an(n-1)+[an+a(n-1)][an-a(n-1)]2[an+a(n-1)]=[an+a(n-1
∵{log2an}是公差为-1的等差数列∴log2an=log2a1-n+1∴an=2log2a1−n+1=a1•2−n+1∴S6=a1(1+12+…+132)=a1•1−1261−12=38,∴a1
(1)a1=(a1+1)24,解得a1=1,当n≥2时,由an=Sn-Sn-1=(an+1)2−(an−1+1)24,得(an-an-1-2)(an+an-1)=0,又an>0,所以an-an-1=2
(1)当n=1时,a1=s1=14a21+12a1−34,解出a1=3,又4Sn=an2+2an-3①当n≥2时4sn-1=an-12+2an-1-3②①-②4an=an2-an-12+2(an-an
4sn=(an+1)^24S(n-1)=(a(n-1)+1)^24Sn-4S(n-1)=4an=an^2+2an+1-(a(n-1)+1)^2(an-1)^2=(a(n-1)+1)^2an-1=a(n
根号Sn的通项公式是nSn=n^2an=Sn-Sn-1=n^2-(n-1)^2=2n-1
Sn、an、1成等差,则2an=Sn+1(n=1时,得a1=1),当n≥2时,有2a(n-1)=S(n-1)+1,则2an-2a(n-1)=an,即an/[a(n-1)]=2=常数,所以{an}是等比
由题意知2an=Sn+1/2,an>0,当n=1时,2a1=a1+1/2,解得a1=1/2,当n≥2时,Sn=2an-1/2,S(n-1)=2a(n-1)-1/2,两式相减得an=Sn-S(n-1)=