a1=3 an=sn-1 2n
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1/S(n+1)=3/Sn+4令1/Sn=bn则有b(n+1)=3bn+4b(n+1)+2=3(bn+2)等比数列,则bn+2=(b1+2)*3^(n-1)b1=1/S1=1/a1=1所以bn=3^n
取倒数1/(Sn+1)=(4n+3)/Sn令bn=1/(Sn)得b1=1b(n+1)=bn*(4n+3)得b(n+1)/bn=4n+3(1)同理bn/(bn-1)=4(n-1)+3(2)...b2/b
http://zhidao.baidu.com/question/88231937.html?fr=qrl&cid=983&index=2S1=a1=-(2/3),S2+1/S2+2=a2,因为S2=
(Ⅰ):证明:∵Sn=12(n+1)(an+1)−1,∴Sn+1=12(n+2)(an+1+1)−1∴an+1=Sn+1−Sn=12[(n+2)(an+1+1)−(n+1)(an+1)]整理,得nan
S1=a1=-(2/3),S2+1/S2+2=a2,因为S2=(a1+a2),所以S2+1/S2+2=S2-a1=S2+2/3,解得S2=-(3/4),同理,S3+1/S3+2=a3=S3-S2=S3
(1)∵Sn+1=2Sn+3n+1,∴当n≥2时,Sn=2Sn-1+3(n-1)+1,两式相减得an+1=2an+3,从而bn+1=an+1+3=2(an+3)=2bn(n≥2),∵S2=2S1+3+
我们都知道在数列里有怎么一个隐藏的条件:an=sn-s(n-1)所以a(n+1)=s(n+1)-sn因为a(n+1)=3Sn所以S(n+1)/Sn=4即{s(n+1)/sn}是以首项为4,公比为4的等
(1)由2an=Sn*S(n-1),an=Sn-S(n-1)则:2[Sn-S(n-1)]=Sn*S(n-1)2Sn-2S(n-1)=Sn*S(n-1)两边同时除以Sn*S(n-1)2/S(n-1)-2
当n≥时an=sn-s(n-1)于是sn=n(2n-1)[sn-s(n-1)]得(2n+1)(n-1)sn=n(2n-1)s(n-1)变形为[(2n+1)/n]sn=[(2n-1)/(n-1)]s(n
取倒数得:1/a(n+1)=(2an+1)/an=2+1/an;所以1/a(n+1)-1/an=2,又a1=1,那么1/an=2n-1,所以an=1/(2n-1)(1/an是等差数列)当n>1时bn=
Sn+S(n+1)=5(a(n+1))/3因为S(n+1)=SN+A(N+1)所以Sn+SN+A(N+1)=5a(n+1)/32SN=2a(n+1)/3SN=a(n+1)/3S(N-1)=AN/3SN
不是这样的1、A(n+1)=S(n+1)-Sn=Sn+3^n>>>>S(n+1)-3^(n+1)=2Sn+3^n-3^(n+1)=2Sn-2×3^n=2[Sn-3^n]则:[S(n+1)-3^(n+1
因为Sn+Sn-1=3an所以Sn-1+Sn-1+an=3an2Sn-1=2anSn-1=an因为Sn=an+1所以Sn-Sn-1=an+1-anan=an+1-an2an=an+1an+1/an=2
Sn-1=(n-1)(n-1)an-1Sn-Sn-1=an=nnan-(n-1)(n-1)an-1(nn-1)an=(n-1)(n-1)an-1an=(n-1)/(n+1)*(n-2)/(n-1)*…
已知数列a‹n›首相a₁=3,通项a‹n›和前n项和S‹n›之间满足2a‹n›=S̸
an=sn-s(n-1)代入得Sn=2S(n-1)+2^n,即Sn/2^n=S(n-1)/2^(n-1)+1所以Sn=(n+1/2)*2^n,所以an=Sn-S(n-1)=n*2^n+2^(n-1).
(1)An=3(1+2^n)(2)由题知,Sn=2An+3n-12=6(2^n-1)+3nBn=(An-3)/(Sn-3n)(A(n+1)-6)=(3*2^n)/(6(2^n-1))(3(2^(n+1
a(n+1)=1/3Snsn=3a(n+1)s(n-1)=3anan=sn-s(n-1)=3a(n+1)-3ana(n+1)/an=4/3an为首相1公比4/3等比a1,a3,a5,.a2n-1为首相
数列{A(n)},A1=1,A(n+1)=3A(n)+4.求A(n)和S(n).1.A(n+1)=3A(n)+4--->A(n)=3A(n-1)+4==3[3A(n-2)+4]+4==(3^2)A(n
An+1=1/3Sn3An+1=Sn(1)3An=Sn-1(2)(1)-(2)得3An+1=4An(n大于等于2),所以An是以A2为首项q=4/3的等比数列A2=1/3A1,所以A2等于1/3An=