若数列{an]满足前n项和Sn=2an-4,bn+1=an+2bn,且b1=2,求:bn;{bn}的前n项和Tn

Sn=2an-4
an = Sn - S(n-1)

Sn = 2[Sn - S(n-1)] - 4
Sn = 2S(n-1) + 4
Sn + 4 = 2[S(n-1) + 4]
(Sn + 4)/[S(n-1) + 4] = 2

Sn + 4 是一个等比数列, 公比为2
Sn + 4 = (S1 + 4) * 2^(n-1)
Sn = (a1 + 4)*2^(n-1) - 4
(符号 ^ 表示乘方）

an = Sn - S(n-1)
= [(a1 + 4)*2^(n-1) - 4] - [(a1 + 4)*2^(n-2) - 4]
= (a1 + 4)*2^(n-2)

令 n = 1
a1 = (a1 + 4)/2
a1 = 4

an = (4 + 4)*2^(n-2) = 2^(n+1)

b(n+1) = an + 2bn = 2^(n+1) + 2bn

b1 = 2
b2 /2 = 2 + b1
b3 /2^2 = 2 + b2 /2
b4 /2^3 = 2 + b3 /2^2
b5 /2^4 = 2 + b4 /2^3
……
bn /2^(n-1) = 2 + b(n-1) /2^(n-2)

以上各等式相加 ，消去 b1, b2/2, b3/2^3 …… , b(n-1)/2^(n-2)，最后余留为
bn/2^(n-1) = 2 + 2 + 2 + …… + 2 = 2n

bn = n*2^n

Tn = b1 + b2 + b3 + …… + bn
= 2 + 2*2^2 + 3*2^3 + …… (n-1)*2^(n-1) + n*2^n

Tn/2 = 1 + 2*2 + 3*2^2 + 4*2^3 + …… (n-1)*2^(n-2) + n*2(n-1)

Tn - Tn/2
= -1 +