设数列{an}和{bn}满足a1=b1=6,a2=b2=3,且数列{a(n+1)-an}是等差数列,数列{bn-2}是等

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设数列{an}和{bn}满足a1=b1=6,a2=b2=3,且数列{a(n+1)-an}是等差数列,数列{bn-2}是等比数列
1 求数列{an}和{bn}的通项公式
2 是否存在k属于正整数,使ak-bk属于（0,1/2）?若存在,求出k,若不存在,为什么?

1.a2-a1=d=-3
a3-a2=-3
a4-a3=-3
...
an-a(n-1)=-3
叠加
an-a1=-3(n-1)
所以an=-3n+9
b2/b1=3/6=1/2
b3/b2=1/2
...
bn/b(n-1)=1/2
同理,叠乘：
bn/b1=(1/2)∧(n-1)
所以 bn=6(1/2)∧（n-1)
2.不存在
ak-bk=-3n+9-6(1/2)∧（n-1)=f(x)
n=1时 f(x)=0
n=2 f(x)=0
n≥3 f（x）＜0,所以不存在
应该没算错,但是第二问一般都是存在的.你再算算.

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