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

(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

(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)/b...