sn 2an 3n
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数列的前n项和sn=2an - 3n,求通项公式
sn=2an-3n,s(n-1)=2a(n-1)-3(n-1),两式相减,并整理得,an=2a(n-1)+3 由an=2a(n-1)+3得,an+3=2(a(n-1)+3),所以数列{an+3}是首项为a1+3=...
数列{an}的前n项和Sn=2an - 3n(n∈N)。
解：Sn=2an-3n，S(n-1)=2a(n-1)-3(n-1)，两式相减得an=2a(n-1)+3，由an+X=2[a(n-1)+X]比较上式可得出X=3，使得数列{an+X}为等比数列。
设数列{an}的前n项和为Sn=2n2+3n求数列{an}的通项公式...
解答:解:由Sn=2n2+3n,当n=1时,a1=S1=5;当n≥2时,an=Sn-Sn-1=2n2+3n-[2(n-1)2+3(n-1)]=4n+1.验证n=1时上式成立.∴an=4n+1.
已知数列{an}的前n项和为Sn,2Sn=3an - 3(n∈N*).(1)求...
(an-1)-32(an-1-1),整理得:an=3an-1,又∵2S1=3a1-3,即a1=3,∴数列{an}是首项、公比均为3的等比数列,∴其通项公式an=3n;(2)由(1)可知bn=log3an+an=log33n....
数列{an}前n项和为sn,sn=2an - 3n(n属于n+),求数列{an}的通...
an=2*a(n-1)+3 即an+3=2*(a(n-1)+3)而a1+3=6,所以an+3=6*2^(n-1)即an=6*2^(n-1)-3=3*2^n-3 解:a1=S1=2a1-3,得a1=3Sn=2an-3nS(...
...an}的前n项和为Sn,若对于任意的正整数n都有Sn=2an - 3n
所以bn为等比数列 a1=3,a2=9,a3=21 b1=6,b2=12,b3=24 bn=3*2^n an+3=3*2^n an=3*2^n-3 nan=3n*2^n-3n 前n项和为Sn Sn=3(1*2^1+2*2^2...
等比数列的前n项和的Sn,S2n,S3n有何关系
∴S(2n)=a1+a2+a3+……+an+a(n+1)+a(n+2)+……+a(n+n)=Sn+Sn*q^n=Sn*(1+q^n)∴S(3n)=Sn+Sn*q^n+Sn*q^(2n)=Sn*(1+q^n+q^(2n))2...
已知数列{an}的前n项和为Sn,且2Sn=3an - 1.(1)求数列{an...
(1)n=1时,a1=1.∵2Sn=3an-1,∴2Sn+1=3an+1-1,∴an+1=3an,∴an=3n-1.(2)∵bn=n 3n-1,∴Tn=1 30+2 31+3 32+…+(n-1) 3n-2+n 3n-1,3 Tn=...
(2010大连一模)数列{an}的前n项和为Sn,且Sn=n2+3n...
解答:解:(I)∵sn=n2+3n2∴n≥2时,an=sn sn 1=n2+3n2 (n 1)2+3(n 1)2=n+1∵n=1时,a1=s1=2又∵a1=S1=2也满足上式,∴an=n+1(n∈N*)(II)∵...
数列an的前n项和满足Sn=2an - 3n.求通项公式
a1=2a1-3a1=3Sn=2an－3nSn-1=2an-1-3n+3an=2an-2an-1-3-an=-2an-1-3an=2an-1 +3an+3=2an-1 + 6(an+3)=2(an-1+3)an+3=6×2^(n-1)...