数列an首项a1=1前n项和sn与an之间满足an=2Sn^2 2Sn-1
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设cn=an/2^n,求证cn是等差数列由cn=an/2^n,知an=2^n*cn,且a(n+1)=2^(n+1)*c(n+1),a(n-1)=2^(n-1)*c(n-1),由bn=2an-4a(n-
S(n+1)=2Sn+3n+1则S(n+1)-Sn=Sn+3n+1即a(n+1)=Sn+3n+1所以Sn=a(n+1)-3n-1所以S(n-1)=an-3(n-1)-1用上式减下式:Sn-S(n-1)
(1)因p为常数,a1=1,故当n=1时,2Sn=2a1=2=p(2*1+1-1)=2p,所以,p=1.Sn=n(an+a1)/2=n(an+1)/22Sn=n(an+1)=2an²+an-
(1)数列{an}中,a1=1,前n项和Sn=n+23an,可知S2=43a2,得3(a1+a2)=4a2,解得a2=3a1=3,由S3=53a3,得3(a1+a2+a3)=5a3,解得a3=32(a
(1)an=sn-s(n-1)就有sn-s(n-1)+2sn*s(n-1)=0两边同除以sn*s(n-1)得1/sn-1/s(n-1)=2{1/sn}是等差数列1/sn=1/s1+(n-1)d=2n-
1)由S(n+1)=4an+2,知S(n)=4a(n-1)+2,两者相减,得S(n+1)-S(n)=a(n+1)=4[an-a(n-1)]由bn=a(n+1)-2an知,b(n-1)=an-2a(n-
a(n+1)=2an/(an+1)1/a(n+1)=1/2(1/an+1)1/a(n+1)-1=1/2(1/an-1)[1/a(n+1)-1]/(1/an-1)=1/2(1/a1-1)=3/2-1=1
n=an+1S(n+1)=2Sn+n+5.1Sn=2S(n-1)+n-1+5=2S(n-1)+n+4.2(1)-(2)得S(n+1)-Sn=2[Sn-S(n-1)]+1a(n+1)=2an+1a(n+
(1)an+2Sn·S(n-1)=0(n≥2),又an=Sn-S(n-1)所以Sn-S(n-1)+2Sn·S(n-1)=0(n≥2)两边同时除以Sn·S(n-1),得1/S(n-1)-1/sn+2=0
S(n+1)=4An+2(1)S(n)=4A(n-1)+2(n≥2)(2)(1)-(2)得,A(n+1)=4A(n)-4A(n-1)(n≥2)[A(n+1)-2An]/[A(n)-2A(n-1)]=[
由题意得:2S(n+1)=4Sn+a1,则2Sn=4S(n-1)+a1解得:a(n+1)=2an,则{an}为等比数列,公比q=2所以,an=a1q^(n-1)=2^n同样:2S(n+1)=4Sn+a
2S(n+1)-Sn=22S(n+1)=Sn+22S(n+1)-4=Sn-2[S(n+1)-2]/(Sn-2)=1/2,为定值.S1-2=a1-2=1-2=-1,数列{Sn-2}是以-1为首项,1/2
是a(n+1)=2an/(an+1)吧a(n+1)=2an/(an+1)1/a(n+1)=(an+1)/(2an)=(1/2)(1/an)+(1/2)1/a(n+1)-1=(1/2)(1/an)-(1
an=Sn-Sn-1=-SnS(n-1)(Sn-Sn-1)/[SnS(n-1)]=-11/S(n-1)-1/Sn=-11/Sn-1/S(n-1)=1,为定值.1/S1=1/a1=1/(1/2)=2数列
a(n+1)=s(n)+3(n+1);an=s(n-1)+3n;两边同减a(n+1)-an=s(n)-s(n-1)+3=an+3所以a(n+1)=2*an+3bn=an+3an=bn-3a(n+1)=
(1)∵a[n]=S[n-1]+2(n>=2)∴a[n+1]=S[n]+2将上述两式相减,得:a[n+1]-a[n]=a[n]即:a[n+1]/a[n]=2∵a[1]=2∴a[n]是首项和公比都是2的
(1)S(n)-S(n-1)=an=2S(n-1)+3^n,即S(n)=3S(n-1)+3^n,两边同时除以3^n,得S(n)/(3^n)=S(n-1)/[3^(n-1)]+1,当n=1时,S(1)/
因为a(n+1)=S(n+1)-S(n)=S(n)+3n+1即a(n+1)=S(n)+3n+1(1)所以a(n)=S(n-1)+3(n-1)+1(2)(1)-(2)得a(n+1)-a(n)=S(n)-
a1=4>0,n≥2时,an的表达式为两算术平方根之和的一半,又算术平方根恒非负,因此{an}各项均非负,√Sn恒有意义.n≥2时,an=Sn-S(n-1)=[√Sn+√S(n-1)]/2[√Sn+√
由2an-1=√1+8s(n-1)平方得an^2-an=2S(n-1)所以a(n-1)^2-a(n-1)=2S(n-2)^2两式相减整理得[an+a(n-1)][an-a(n-1)-1]=0因为an>