色谱柱货号是S N还是B N?

∵Sn＋bn＝2Sn－Sn-1＝bn∴2Sn－Sn-1＝2∴2(Sn－2)＝Sn-1－2∴﹛Sn－2﹜是等比数列∴Sn－2＝（S1－2）×（1/2）^(n－1)∵Sn＋bn＝2∴S1＋b1＝2b1＝2

1设an工笔qbn-bn-1=log3an-log3an-1=log3(an/an-1)=log3q=d所以bn为等差数列2b1=log3a1=4由题意可知d

a(n)=81*q^(n-1),b(n)=log_{3}[a(n)]=log_{3}[81*q^(n-1)]=log_{3}(81)+log_{3}[q^(n-1)]=4+(n-1)log_{3}(q

1.n=1时,a1+S1=2a1=1a1=1/2n≥2时,Sn=1-anS(n-1)=1-a(n-1)Sn-S(n-1)=an=1-an-1+a(n-1)2an=a(n-1)an/a(n-1)=1/2

(1)等比数列｛an},首项为81设an=a1*q^(n-1)=81*q^(n-1)数列{bn}满足bn=log3为底an∴bn=log3为底[81*q^(n-1)]=log3为底81+log3为底q

属于物理,不过化学研究上运用色谱

1an=Sn-Sn-1=4an-4an-14an-1=3anan/an-1=4/3a1=4a1-3,a1=1an=1*(4/3)^(n-1)2b1=2b2=a1+b1=3b3=b2+a2=2+1+(4

根据Sn=2an-1与s（n-1）=2a（m-1）-1两式相减,得an/a(n-1)=2,即an是2为公比的等比数列.a1=2a1-1,得a1=1所以an的通项公式为an=2^（n-1）所以bn+1=

Sn=1/2(Bn+1/Bn)而S(n-1)=Sn-Bn=1/2(1/Bn-Bn)所以Sn+S(n-1)=1/Bn以及Sn-S(n-1)=BnSn^2-S(n-1)^2=1而S1=a1=1/2(B1+

n≥2时,有bn=Sn-SSn=(bn+n/bn)/2=(Sn-S+n/(Sn-S))/2故有Sn^2-S^2=n(n≥2)n=1时,有b1=S1=(b1+1/b1)/2,b1=1或者-1(舍去负值)

再答：满意采纳，不懂追问，谢谢

n=n(n+1)=n^2+nSn=b1+b2+...+bn=(1^2+1)+(2^2+2)+...+(n^2+n)=(1^2+2^2+...+n^2)+(1+2+...+n)=n(n+1)(2n+1)

Sn=1-(1/2)bn、S1=b1=1-(1/2)b1,则b1=2/3.b(n+1)=S(n+1)-Sn=(1/2)bn-(1/2)b(n+1),则b(n+1)/bn=1/3.所以,数列{bn}是首

1/n^2+n=1/n(n+1)列项得1/n(n+1)=1/n-1/(n-1)然后累加

GP-goldplatedSG/GP-satingold/goldplatedACP-antiqurcopperplatedABP-antiquebrassplated

平方和的公式为S＝n(n+1)(2n+1)/6所以,Sn＝4×n(n+1)(2n+1)/6＋4×n(n+1)/2＝2n(n+1)(2n+1)/3＋2n(n+1)＝2n(n+1)(2n+4)/3＝4n(

1=24bn=-3n+27≥03n≤27n≤9当n=9时bn}的前n项和Sn值最大b9=-3*9+27=0sn=(b1+b9)*9/2=(24+0)*9/2=54

设{an}公差为d.a7-a3=4d=13-5=8d=2a1=a3-2d=5-4=1an=a1+(n-1)d=1+2(n-1)=2n-1数列{an}的通项公式为.n=1时,S1=b1=2b1-1b1=

Sn+bn=2S(n-1)+b(n-1)=2相减得Sn+bn-S(n-1)-b(n-1)=02bn=b(n-1)bn/(bn-1)=1/2由上式可知是等比数列,公比是1/2bn=b1q^(n-1)由S

已知等比数列an,首项为81,数列bn满足bn=log3an,其前n项和sn(1)证明：bn-b(n-1)=log(3)an-log(3)an-1=log（3）an/a(n-1)=log(3)q∵b1