已知数列an首项为2且对任意n属于n*,都有1 a1a2
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(1)由已知有:2a1=4096得a1=2048,又an+sn=4096,an+1+Sn+1=4096,两式相减得an+1=an/2,所以an是以1/2为公比的等比数列,故an=2048*(1/2)^
(1)由a1+S1=1及a1=S1得a1=12.又由an+Sn=n及an+1+Sn+1=n+1,得an+1-an+an+1=1,∴2an+1=an+1.∴2(an+1-1)=an-1,即2bn+1=b
2倍的根号下Sn=An+1根号下Sn=(An+1)/2Sn=(An+1)^2/4An=Sn-S(n-1)=(An+1)^2/4-(A(n-1)+1)^2/4即:4An=(An)^2+2An-[A(n-
两边平方,得(an+2)^2/4=2Sn,两边同时除2,得Sn=(an+2)^2/8,S_(n+1)-Sn=a_(n+1)=[(a_(n+1)+2)^2-(an+2)^2]/8,完全平方式化成三项式后
无事推导下,宝刀未老!
(1)Sn+n=2anSn=2an-nS(n-1)=2a(n-1)-(n-1)an=Sn-S(n-1)=[2an-n]-{2a(n-1)-(n-1)}=2an-2a(n-1)-1an=2a(n-1)-
(1)an+Sn=na(n+1)+S(n+1)=n+1两式相减2a(n+1)-an=1,即2(a(n+1)-1)=an-1,2b(n+1)=bn而a1+a1=1,a1=1/2,b1=-1/2,{bn}
当n=1时,有a2/a1=(4*1-1)/(2*1-1)=3,∴a2=3a{an}不是等差数列吗?那好,公差d=a2-a1=2a∴an=a1+(n-1)*d=a*(2n-1),n∈N*再问:谢谢了,还
(1)证明:∵Sr/St=(r/t)² 对于r=n,t=1时同样成立 S(n)/S(1)=n^2,S(n)=n^2S(1)=n^2a(1),S(n+1)=(n+1)^2a(1),a(n+
2√Sn=an+1则有,4Sn=(an+1)²4a(n+1)=4[S(n+1)-Sn]=[a(n+1)+1]²-(an+1)²=[a(n+1)]²+2a(n+1
由4Sn=(an+1)^2得4S(n+1)=(a(n+1)+1)^2两式相减4a(n+1)=[a(n+1)+an+2]*[a(n+1)-an]化简2(a(n+1)+an)=(a(n+1)+an)(a(
an+sn=na(n+1)+s(n+1)=n+1a(n+1)-an+a(n+1)=1a(n+1)-1=0.5(an-1)即{an-1}是以a1-1=-0.5为首项0.5为公比的等比数列
1、an=Sn-S(n-1)所以2Sn-S(n-1)=20482Sn=S(n-1)+20482Sn-4096=S(n-1)+2048-40962(Sn-2048)=S(n-1)-2048(Sn-204
2an=Sn+n2a(n-1)=S(n-1)+n-1相减2an-2a(n-1)=an+1an=2a(n-1)+1同时加1an+1=2[a(n-1)+1][an+1]/[a(n-1)+1]=2是等比数列
(1)2an=n+Sn2a(n+1)=n+1+S(n+1)相减得2【a(n+1)-an】=1+a(n+1)a(n+1)=2an+1b(n+1)=a(n+1)+1=2(an+1)=2bna1=1an=2
(1)令Cn=an*a(n+1)=2^n,则C(n+1)=a(n+1)*a(n+2)=2^(n+1),两式相除有2=a(n+2)/an即{a(n+2)/an}是以2为公比的等比数列由a1=1易得a2=
由题意得1/a1a2+1/a2a3…1/anan-1=(n-1)/a1an①原式-①得1/anan+1=n/a1an+1-(n-1)a1an整理得2=nan-(n-1)an+1两边同时除以n(n-1)
(1)证明:∵n、an、Sn成等差数列∴2an=n+Sn,∴2(Sn-Sn-1)=n+Sn,∴Sn+n+2=2[Sn-1+(n-1)+2]∴Sn+n+2Sn−1+(n−1)+2=2∴{Sn+n+2}成
(1)∵an+Sn=4096,∴a1+S1=4096,a1=2048.当n≥2时,an=Sn-Sn-1=(4096-an)-(4096-an-1)=an-1-an∴anan−1=12an=2048(1
a20=a5+15d,a20=2a5,所以2a5=a5+15d故a5=15d又a5=a1+4d,故a1=11d.s5=5(a1+a5)\2=5×26d\2=105故d=21\13