Question

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

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。

麻烦告诉解答过程

Answer 1

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

追问

谢谢

追答

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

Answer 2

楼上答案为错解，因为此题所求d为3，虽然公差有误，但思路还是正确的，即运用错位相减法求解