已知数列AN的首项为2,且对任意的N属于N,都有A1A2分支1
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n=1时,S1=a1=2a1-1,a1=1n≥2时,an=Sn-S(n-1)=(2an-1)-(2a(n-1)-1)an=2a(n-1),故an=2^(n-1).
可以用an与Sn之间的关系求当n》2时an=Sn-S(n-1)=2an-2a(n-1)即an=2a(n-1)即数列{an}是等比数列当n=1时a1=S1=2a1-1a1=1an=2的n-1次方
因为An+1=2SnAn=2S(n-1)所以A(n+1)-An=2AnA(n+1)/An=3是公比为3,首项a1=1的等比数列,An=A1*q^(n-1)即An=3^(n-1)
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-
先列式4*(S1)=(a1)*(a2).14*(S2)=(a2)*(a3).2...4*(Sn)=(an)*(a(n+1)).n2式-1式,3式-2式,.可以得出a3-a1=4a4-a2=4...an
(1)设bn=log2(an+1),则{bn}为等差数列,又a1=1,a3=7,所以b1=log2(1+1)=1,b2=log(7+1)=3,所以公差d=1.所以bn=b1+(n-1)d=1+(n-1
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楼貌似错了!(a1^2-3a1=6a1与An^2+3An=6Sn矛盾)An^2+3An=6SnA(n+1)^2+3A(n+1)=6S(n+1)后减前得A(n+1)^2+3A(n+1)-An^2-3A
1.a(n+1)-an=1,为定值,又a1=1,数列{an}是以1为首项,1为公差的等差数列.an=1+n-1=nn=1时,S1+b1=2b1=2b1=1n≥2时,Sn=2-bnS(n-1)=2-b(
(1)由题意知an=1/2(3n+Sn)对一切正整数n恒成立,又当n=1时,s1=a1.所以a1=1/2(3+a1),所以a1=3(2)证明:由题意知an=1/2(3n+Sn)对一切正整数n恒成立,即
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(
Sn=n^2+an/2S(n-1)=(n-1)^2+a(n-1)/2an=Sn-S(n-1)=2n-1+[an-a(n-1)]/2an=4n-2-a(n-1)an-2n=-a(n-1)+2n-2an-
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(
Sn=an-2;Sn-1=an-3;an=an-2-an-3;条件不足,a1,a2没有初始值吗
Sn+an=n^2+3n+5/2①当n=1时,S1+a1=1^2+3*1+5/2=13/2而S1=a1,所以2a1=13/2,即a1=13/4,所以a1-1=9/4;又S(n-1)+a(n-1)=(n
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
(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
3an=3+2Sn3a(n-1)=3+2S(n-1)相减,有:an=3a(n-1),即等比数列,又由3a1=3+2S1,可得a1=3,则an=3^n
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