Question

数列求解。。。

Answer 1

a(n)>0.
6s(n) = [a(n)]^2 + 3a(n) + 2,
6a(1) = 6s(1) = [a(1)]^2 + 3a(1) + 2,
0 = [a(1)]^2 - 3a(1) + 2 = [a(1)-2][a(1)-1],
a(1)=1或a(1)=2.

6s(n+1) = [a(n+1)]^2 + 3a(n+1) + 2,
6a(n+1) = 6s(n+1)-6s(n) = [a(n+1)]^2 + 3a(n+1) - [a(n)]^2 - 3a(n),
0 = [a(n+1)]^2 - [a(n)]^2 - 3a(n+1) - 3a(n) = [a(n+1)+a(n)][a(n+1)-a(n) - 3],
因 a(n+1)+a(n)>0.
所以
0 = a(n+1)-a(n)-3,
a(n+1) = a(n)+3,
{a(n)}是首项为a(1)=1或a(1)=2,公差为3的等差数列 。
a(n) = 1 + 3(n-1) = 3n-2
或 a(n) = 2 + 3(n-1) = 3n-1.

若 a(n) = 3n-2,则a(1)=1,a(2)=4,a(3)=7. 4^2 - 1*7 = 16 - 7 = 9 = [a(2)]^2 - a(1)a(3),与 a(1),a(2),a(3)是等比数列的前三项矛盾 。
因此，只能有，
a(n) = 3n-1.
验证如下：a(1)= 2,a(2)=5,a(3)=8.
啊，也矛盾啊 。[a(2)]^2 - a(1)a(3) = 5^2 - 2*8 = 25-16 =9,也与a(1),a(2),a(3)是等比数列的前三项矛盾。
楼主。。。。题目有问题吧？？？

追问

我也觉得，你看答案

追答

看解答，应该是a(1),a(2),a(6)是等比数列b(n)的前三项。

追问

可题目。。

追答

题目属于，印刷错误，原谅他们吧，楼主。。他们不知道自己在干啥。。

追问

好吧。。

追答

接前。。。
a(n) = 1 + 3(n-1) = 3n-2
或 a(n) = 2 + 3(n-1) = 3n-1.

若 a(n) = 3n-1,则a(1)=2,a(2)=5,a(6)=17.与 a(1),a(2),a(6)是等比数列的前三项矛盾 。
若a(n) = 3n-2.a(1)= 1,a(2)=4,a(6)=16.满足[a(2)]^2 = a(1)*a(6).
这样，
a(n) = 3n-2.

b(1) = a(1) = 1,
b(2) = a(2) = 4,
b(n) = 4^(n-1).

1<=k<= n时，a(k)b(n-k+1) = (3k-2)4^(n-k) = (3k-2)4^(-k)*4^n.

t(n) = a(1)b(n) + a(2)b(n-1) + ... + a(n-1)b(2) + a(n)b(1)

= 1*4^(n-1) + 4*4^(n-2) + ... + (3n-5)4^1 + (3n-2)
= 4^n*[(3*1-2)(1/4) + (3*2-2)(1/4)^2 + (3*3-2)(1/4)^3 + ... + (3n-5)(1/4)^(n-1) + (3n-2)(1/4)^n],

4t(n) = 4^n[(3*1-2) + (3*2-2)(1/4) + (3*3-2)(1/4)^2 + ... + (3n-5)(1/4)^(n-2) + (3n-2)(1/4)^(n-1)]
3t(n) = 4t(n) - t(n)
= 4^n[ 3-2 + 3(1/4) + 3(1/4)^2 + ... + 3(1/4)^(n-1) - (3n-2)(1/4)^n]
= 3*4^n[1 + 1/4 + (1/4)^2 + ... + (1/4)^(n-1)] - 4^n[2+(3n-2)(1/4)^n]
= 3*4^n[ 1 - (1/4)^n]/(1-1/4) - 2*4^n - (3n-2)
= 4*4^n[1 - (1/4)^n] - 2*4^n - (3n-2)
= 2*4^n - 4 - (3n-2)
= 2*4^n - 3n - 2,

3t(n) + 1 = 2*4^n - 3n - 1, a(n+1) = 3n+1, b(n+1) = 4^n, 3t(n)+1=2b(n+1)-a(n+1),符合题意。

追问

嗯，谢谢。那个如果没错，这题其实不难