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

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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
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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
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

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 + 2*(1-2) + 2^2 * (2-3) + 2^3 *(3-4) + …… + 2^(n-1) * [(n-1) - n] + n*2^n
= - [1 + 2 + 2^2 + 2^3 + …… + 2^(n-1)] + n*2^n
= - (2^n -1) + n*2^n
= (n-1)*2^n + 1
Tn = (n-1)*2^(n+1) + 2
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附录：本题目比较庞大麻烦.所以不要轻易关闭问题.有看不懂的地方请指出,我再补充解释.谢谢.

对楼上结果的检验：
a1 = 4
a2 = 8
S2 = 4 + 8 = 12 = 2*8 -4
a3 = 16
S3 = 4 + 8 + 16 = 28 = 2*16 - 4
b1 = 2
b2 = a1 + 2b1 = 4 + 4 = 8
b3 = a2 + 2b2 = 8 + 2*8 = 24
b4 = a3 + 3b...

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对楼上结果的检验：
a1 = 4
a2 = 8
S2 = 4 + 8 = 12 = 2*8 -4
a3 = 16
S3 = 4 + 8 + 16 = 28 = 2*16 - 4
b1 = 2
b2 = a1 + 2b1 = 4 + 4 = 8
b3 = a2 + 2b2 = 8 + 2*8 = 24
b4 = a3 + 3b3 = 16 + 2*24 = 64
T1 = 2
T2 = 2+8 = 10
T3 = 10 + 24 = 34
T4 = 34 + 64 = 98
按照楼上的公式
Tn = (n-1)*2^(n+1) + 2
T1 = 2
T2 = (2-1)*2^(2+1) + 2 = 8 + 2 = 10
T3= (3-1)*2^(3+1) + 2 = 32 + 2 = 34
T4 = (4-1)*2^(4+1) + 2 = 3*32 + 2 = 98
都成立啊

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