数列an为各项均为正数的等比数列,它的前n项和为80
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a1=3,所以S2=6+d,S3=9+3db1=1,b2=q,b3=q^2所以(6+d)q=64(9+3d)q^2=960相除(9+3d)/(6+d)*q=15q=15(6+d)/(9+3d)代入(6
请把题目拍照上传.我见过类似的题说是an为等比数列.(本人每天白天在线)再问:发了图片快看一看再答:我想问一下你是高几的,数学归纳法学了没有。可以根据已知条件依次求出a1=1,a2=1/2,a3=1/
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
①依题意,得(an+2)/2=根号下(2Sn),∴a1+2=2根号下(2S1)=2根号下(2a1),∴(a1-2)的平方=0,∴a1=2,由a2+2=2根号下(2S2)=2根号下[2(2+a2)],得
∵(an+1)²-an+1×an-2an²=0∴(an+1+an)(an+1-2an)=0∴an+1-2an=0,an+1+an=0(舍去)∴an+1=2an∴an是等比数列,设a
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)=
等比数列,则:a1a3=(a2)²,a3a5=(a4)²,则:a1a3+2a2a4+a3a5=(a2)²+2a2a4+(a4)²=(a2+a4)²=1
sn=(1/8)(an+2)²S(n-1)=(1/8)[a(n-1)+2]²an=Sn-S(n-1)=(1/8){(an+2)²-[a(n-1)+2]²}=(1
6Sn=an^2+3an+26S(n-1)=a(n-1)^2+3a(n-1)+26Sn-6S(n-1)=6an=an^2+3an+2-a(n-1)^2-3a(n-1)-26an=an^2+3an-a(
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
由题意得1S3=a1+a2+a3=7……1;6a2=a1+1+a3+6……22式+1式得a2=2……3将3式代入12得q=2或1/2a1=4或1an=4*(1/2)^(n-1)或an=2^(n-1)2
(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
a1+a2+...+an=(1/2)(an²+an)a1+a2+...+a(n-1)=(1/2)(a(n-1)²+a(n-1))两式相减得an=(1/2)(an²+an)
1.A(n+1)^2*An+A(n+1)*An^2+A(n+1)^2-An^2=0两边同除以A(n+1)²An²1/An+1/A(n+1)+1/An²-1/A(n+1)&
n>=2时,S[n]=1/4*(a[n]+1)^2;S[n-1]=1/4*(a[n-1]+1)^2两式相减得到a[n]=1/4*(a[n]^2+2a[n]-a[n-1]^2-2a[n-1])化简得到a
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)=