在数列an中an>0 sn是它的前n项和 且2根号下sn=an 1
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(1)Sn+1=Sn+an+1=4an-1+2+an+1∴4an+2=4an-1+2+an+1∴an+1-2an=2(an-2an-1)即:bnbn−1=an+1−2anan−2an−1=2 
设公比为q,当q=-1时,等比数列{an}的各项是a,-a,a,-a,a,-a…的形式,a≠0.又已知Sn是实数等比数列{an}前n项和,故当n为偶数时,Sn=0,当n为奇数时,Sn=a,故选D.
1∵an+1=1/2(an-an+1)∴3/2an+1=1/2an∴an+1/an=1/3∴{an}是等比数列∴an=2/27×(1/3)^(n-4)=2×(1/3)^(n-1)2所以a1=2前n项和
S[1]=a[1]=1/2(a[1]+1/a[1]),于是:a[1]=1=√1-√0S[2]=a[2]+1=1/2(a[2]+1/a[2]),于是:a[2]=√2-1,S[2]=√2S[3]=a[3]
∵2√Sn=an+1,∴Sn=(an+1)^2/4∴S(n-1)=(a(n-1)+1)^2/4两式相减,得到an=Sn-S(n-1)=1/4*(an^2-a(n-1)^2)+1/2*(an-a(n-1
是等差数列,首项a1=10,公差是-1/2,通项an=10-(n-1)/2,前n项和Sn=n(a1+an)/2=n(10+10-(n-1)/2)/2=[21^2-1-(n-21)^2]/4,当n=21
an=Sn-Sn-1=4n+1(n>=2),a1=2*1+3=5,满足上式,an通项就是4n+1,即证实等差数列
【条件1:Sn有上界】是【条件2:An收敛】的必要非充分条件.因为An收敛,则An【单调】有界.那么Sn就一定有界.但Sn有界并不能保证An一定【单调】有界即收敛.所以前者应该是后者的必要非充分条件.
an=1/n(n+1)(n+2)=[1/n(n+1)-1/(n+1)(n+2)]/2,a1=1/6所以S1=a1=1/6n>=2时,Sn=a1+a2+...+an=[1/1*2-1/2*3]/2+[1
An=3S(n-1).用原式减去,得A(n+1)-An=3An.A(n+1)=4An.则An为等比数列.
Sn=n^2/(3n+2)Sn-1=(n-1)^2/(3n-1)an=Sn-Sn-1=(3·n^2+n-2)/(9·n^2+3n-2)所以,当n接近正无穷时liman=1/3
(1)an是Sn与2的等差中项即a1=2sn=2an-2所以s(n-1)=2a(n-1)-2an=sn-s(n-1)=2a(n-1)所以an为等比数列公比为2首项为2则an=2^n而点P(bn,bn+
an,Sn,Sn-1/2成等比数列an(Sn-1/2)=Sn^2a2(S2-1/2)=S2^2a2(a2+1/2)=(a2+1)^2a2=-2/3a3(S3-1/2)=S3^2a3(a3-1/6)=(
n+Sn=2an,所以1+s1=2a1=2s1即s1=a1=1且n+1+S(n+1)=2a(n+1)相减得1+a(n+1)=2a(n+1)-2ana(n+1)=2an+1a(n+1)+1=2an+2=
因为6Sn=(an+1)(an+2)(1)所以6Sn-1=(an-1+1)(an-1+2)(2)(1)-(2)则an-an-1=3所以an是等差数列因为6Sn=(an+1)(an+2)可知S1=a1=
因为S10=a1+a2+…+a10,S22=a1+a2+…+a22,又S10=S22,所以a11+a12+…+a22=0,所以12(a11+a22)2=0,即a11+a22=2a1+31d=0,又a1
an+2-3an+1+2an=0an+2-an+1=2(an+1-an){an-an-1}为公比为2的等比数列!an-a(n-1)=2^(n-2)*(a2-a1)=3*2^(n-2)a2-a1=3a3
解an是等差a3=a1+2d=0(1)s4=4a1+6d=-4∴2a1+3d=-2(2)(1)×2-(2)得d=2∴a1=-4∴an=-4+(n-1)×2=2n-6
1.n=1时,S1=a1=(a1²+a1)/2,整理,得a1²-a1=0a1(a1-1)=0a1=0(与已知不符,舍去)或a1=1S1=a1=1n≥2时,Sn=(an²+
Sn=n-2an,Sn-1=(n-1)-2an-1(n大于1)做差an=1-2an-2an-13an-3=2an-1-2(an-1)/[a(n-1)-1]=2/3是常数,经检验,a1=1/3,a2=5