设数列{An}的前项n和为Sn,若对于任意的正整数n都有Sn=2an-3n.设bn=an+3 (1)

设数列{An}的前项n和为Sn,若对于任意的正整数n都有Sn=2an-3n.设bn=an+3 (1)

(1)Sn = 2an-3nn=1,a1= 3an = Sn - S(n-1)= 2an - 2a(n-1) -3an = 2a(n-1) +3an+3 =2( a(n-1) + 3){ an +3 }是等比数列,q=2bn = an+3 是等比数列,q=2(2)an+3 =2( a(n-1) + 3)=2^(n-1) .(a1+3)= 3.2^nan = -3 +3.2^n(3)letS = 1.2^1+2.2^2+...+n.2^n (1)2S = 1.2^2+2.2^3+...+n.2^(n+1) (2)(2)-(1)S = n.2^(n+1) -(2+2^2+...+2^n)=n.2^(n+1) -2(2^n-1)cn = n.an= n(-3 +3.2^n)= 3(n.2^n) - 3nTn = c1+c2+...+cn=3S - 3n(n+1)/2=3n.2^(n+1) -6(2^n-1) - 3n(n+1)/2= 6-[3n(n+1)/2] +(6n-6).2^n======以下答案可供参考======供参考答案1:(1)证明：由Sn=2an-3n得a1=3，S(n+1)=2a(n+1)-3(n+1) S(n+1)-Sn=[2a(n+1)-3(n+1)]-(2an-3n) 即a(n+1)=2[a(n+1)-an]-3 a(n+1)=2an+3 b(n+1)/bn=[a(n+1)+3]/(an+3)=2(an+3)/(an+3)=2 所以数列{bn}是以b1=6为首项，q=2为公比的等比数列 所以bn=b1*q^(n-1)=6*2^(n-1)

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