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数列{an}的前n项和为Sn=2^n+1-2、数列{bn}是首项为a1、公差为d(d≠0)的等差数列、且b1、b3、b1

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数列{an}的前n项和为Sn=2^n+1-2、数列{bn}是首项为a1、公差为d(d≠0)的等差数列、且b1、b3、b11成等比数列
问(1)求数列{an}与{bn}的通项公式(2)设Cn=bn/an、求数列{cn}的前n项Tn.
因为 Sn = 2^(n+1) - 2,则
An = Sn - S(n-1) = [2^(n+1) -2] - (2^n -2) = 2^(n+1) - 2^n = 2^n *(2 - 1) = 2^n
所以,{Bn} 有:
B1 = A1 = 2,B3 = A1 + 2d = 2 + 2d,B11 = A1 + 10d = 2 + 10d
因为 B1、B3、B11 成等比数列,所以有:
(B3)^2 = (2 + 2d)^2 = B1 * B11 = 2 * (2 + 10d)
化简后得到:
4 + 8d + 4d^2 = 4 + 20d,d^2 - 8d = d*(d-8) = 0
所以,d = 8
所以,Bn = A1 + (n-1)*d = 2 + 8 *(n-1) = 8n -6
Cn = Bn/An = (8n - 6)/2^n
Tn = 8*∑n/2^n - 6*∑1/2^n
因为:2* ∑n/2^n - ∑n/2^n = ∑n/2^n
=[1/2^0 + 2/2^1 + 3/2^2 +……+ n/2^(n-1)] - [1/2^1 + 2/2^2 +……+ (n-1)/2^(n-1) + n/2^n]
=1/2^0 + 1/2^1 + 1/2^2 +……+ 1/2^(n-1) + n/2^n
= n/2^n + [(1/2)^0 - (1/2)^n]/(1-1/2)
= n/2^n + 2 - 2 * (1/2)^n
= (n-2)/2^n + 2
∑1/2^n = [(1/2) - (1/2)^(n+1)]/(1-1/2)
= 1 - (1/2)^n
所以,
Tn = 8* (n-2)/2^n + 16 - 6 + 6/2^n
= (8n - 16 + 6)/2^n + 10
= (8n -10)/2^n + 10 再答: 化简后得到: 4 + 8d + 4d^2 = 4 + 20d, d^2 - 3d = d*(d-3) = 0 所以,d = 3 所以,Bn = A1 + (n-1)*d = 2 + 3 *(n-1) = 3n -1 Cn = Bn/An = (3n - 1)/2^n Tn = 3*∑n/2^n - 1*∑1/2^n 因为:2* ∑n/2^n - ∑n/2^n = ∑n/2^n =[1/2^0 + 2/2^1 + 3/2^2 +……+ n/2^(n-1)] - [1/2^1 + 2/2^2 +……+ (n-1)/2^(n-1) + n/2^n] =1/2^0 + 1/2^1 + 1/2^2 +……+ 1/2^(n-1) - n/2^n = -n/2^n + [(1/2)^0 - (1/2)^n]/(1-1/2) = -n/2^n + 2 - 2 * (1/2)^n = 2 - (n+2)/2^n ∑1/2^n = [(1/2) - (1/2)^(n+1)]/(1-1/2) = 1 - (1/2)^n 所以, Tn = 6 - 3* (n+2)/2^n - 1 + 1/2^n = 5 - (3n + 6 - 1)/2^n = 5 - (3n +5)/2^n