已知数列an的各项为正数,4sn=an^2 2an-3
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(1)∵2Sn=an2+n-4(n∈N*).∴2Sn+1=an+12+n+1-4.两式相减得2Sn+1-2Sn=an+12+n+1-4-(an2+n-4),即2an+1=an+12-an2+1,则an
(1)已知a3=4S3=a1+a2+a3---->a1+a2=7-4=3a2*a2=a1*a3------>4a1=a2*a2由1.2可求得a2=2或者a=-6题目已知数列{an}是各项都是正数的等比
a1=2,a2=1,等比1/2,an=2×(1/2)^(n-1).a1=2a2=1,a1=1,a2=1/2,等比1/2,an=1×(1/2)^(n-1).
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
证明:∵Sn=an(an+1)2∴S1=a1(1+a1)2∴a1=1…(1分)由2Sn=a2n+an2Sn-1=a2n-1+an-1⇒2an=2(Sn-Sn-1)=a2n-a2n-1+an-an-1…
∵(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)=
4a(1)=[a(1)+1]^2a(1)=14a(n+1)=[a(n+1)+1]^2-[a(n)+1]^2[a(n)+1]^2=[a(n+1)-1]^2若a(n+1)>1a(n+1)=a(n)+2a(
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
(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
4a(1)=[a(1)+1]^2a(1)=14a(n+1)=[a(n+1)+1]^2-[a(n)+1]^2[a(n)+1]^2=[a(n+1)-1]^2若a(n+1)>1a(n+1)=a(n)+2a(
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