www.1862.net > 在数列{An}中,已知A1=%1,An+1=Sn+3n%1,其中Sn...

在数列{An}中,已知A1=%1,An+1=Sn+3n%1,其中Sn...

解:由an+1=Sn+3n-1(n∈N*)①得an=Sn-1+3n-4(n≥2)②①-②得an+1=2an+3(n≥2)∴an+1+3=2(an+3)(n≥2)又由②得 a2=S1+6-4=a1+2=1∴a2+3=4∴a2+3=2(a1+3)∴an+1+3=2(an+3)(n≥1)∴数列{an+3}是首项为2,公比为2的等比数列∴an+3=2*2n-1=2n∴数列{an}的 an=2n-3(n≥1)

Sn=A(n+1)-3n+1Sn+1-Sn=A(n+2)-A(n+1)-3=A(n+1)A(n+2)=2A(n+1)+3令A(n+2)+λ=2[A(n+1)+λ]则A(n+2)=2A(n+1)+λ得λ=3所以A(n+2)+3=2[A(n+1)+3]所以{An +3}为等比数列.首项为A1+3=2,公比为2.那么An+3=(A

a(n+1)-an=3n-1当n≥2时,an-a(n-1)=3(n-1)-1=3n-4an=[an-a(n-1)]+[a(n-1)-a(n-2)]+……+(a3-a2)+(a2-a1)+a1代入计算即可等到答案

证明:A(n+1)=Sn+3n+1,则An=S(n-1)+3n-2两式想减得A(n+1)-An=Sn+3n+1-(S(n-1)+3n-2)=An+3即A(n+1)+3=2(An+3)即(A(n+1)+3)/(An+3)=2又a1=1 A1+3=4即{An+3}是首项为4,公比为2的等比数列

1.a(n+1)=(1/3)SnS(n+1)-Sn=(1/3)SnS(n+1)=(4/3)SnS(n+1)/Sn=4/3,为定值.S1=a1=1数列{Sn}是以1为首项,4/3为公比的等比数列.Sn=1*(4/3)^(n-1)=(4/3)^(n-1)n≥2时,an=Sn-S(n-1)=(4/3)^(n-1) -(4/3)^(n-2)=4^(n-2)/ 3^(n-1)n=1时,a1=4^(1-2)

解:依题意得:sn-sn-1=an;1/sn-1/sn-1=(sn-1 - sn ) / (snsn-1) =-an/an=-1;s1=a1=2/9;所以1/sn=1/s1+(n-1)*(-1)=11/2-n;所以sn=2/(11-2n);所以an=sn-sn-1=4/[(2n-11)(2n-13)];所以数列{1/sn}是公差为-1,首项为2/9的等差数列. 如有疑问欢迎追问,谢谢采纳.

怀疑你输错了,An+1=3(根号(Sn+1)+根号(sn))Sn+1-Sn=3(√(Sn+1)+√Sn)√(Sn+1)-√Sn=3√Sn是等差数列,首项为1,公差为3所以 √Sn=1+3(n-1)=3n-2Sn=(3n-2)n≥2an=Sn-S(n-1)=(3n-2)-(3n-5)=18n-21 所以 an= 1 n=1 18n-21 n≥2

解 因为an+1=sn+3n 所以an=s<n-1>+3(n-1) 两式相减可得 a<n+1>- an=sn-s<n-1>+3n-3(n-1) 因为sn-s<n-1>=an 所以a<n+1>- an=an+3 ==>a<n+1>=2an+3 ==>a<n+1>+3=2(an+3) 所以数列{an+3}是一个公比为3的等比数列 它的首项为a1=a 所以an+3=a*2^(n-1)==>an=a*2^(n-1)-3

2/[(n+1)n]

解:(一)由题意可知a(n+1)=sn+3n+1 ①,则有an=s(n-1)+3n-2 ②②-①,得 an=Sn-S(n-1)=a(n+1)-an-3,即 a(n+1)=2an+3两边同时除以2^n+1,得 a(n+1)/2^n+1=2an/2^n+1+3/2^n+1(之后用累加..)an/2^n-a(n-1)/2^n-1=3/2^n.

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