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)求证:数列{bn}是

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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)求证:数列{bn}是等比列.
(1)求证:数列{bn}是等比列.(2)求出{An}的通项公式.(3)求数列{NAn}的前n项和.
我想知道这题不是bn+1/bn=2.为什么bn=6*2^(n-1),不是应该bn=6*2^n吗?

(1)
Sn = 2an-3n
n=1,a1= 3
an = Sn - S(n-1)
= 2an - 2a(n-1) -3
an = 2a(n-1) +3
an+3 =2( a(n-1) + 3)
{ an +3 }是等比数列,q=2
bn = an+3 是等比数列,q=2
(2)
an+3 =2( a(n-1) + 3)
=2^(n-1) .(a1+3)
= 3.2^n
an = -3 +3.2^n
(3)
let
S = 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) - 3n
Tn = 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

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