设数列An,Bn满足a1=b1=6,a2=b2=4,a3=b3=3,且数列A(n+1)-An(n属于正整数)是等差数列.

设数列An,Bn满足a1=b1=6,a2=b2=4,a3=b3=3,且数列A(n+1)-An(n属于正整数)是等差数列.
设数列An,Bn满足a1=b1=6,a2=b2=4,a3=b3=3,且数列A(n+1)-An(n属于正整数)是等差数列,sn为数列｛BN}的前几项和,且sn=2n-bn+101）求数列An,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；若不存在说明理由

a(n+1)-a(n)=a+(n-1)d
a=a(2)-a(1)=4-6=-2
a+d=a(3)-a(2)=3-4=-1
d=-1-a=1
a(n+1)-a(n)=-2+n-1=n-3
a(n+1)-(1/2)(n+1)^2 -a(n) + (1/2)n^2 = a(n+1)-a(n)-(2n+1)/2=n-3-(2n+1)/2=-7/2
{a(n)-(1/2)n^2}是首项为a(1)-1/2=11/2,公差为(-7/2)的等差数列.
a(n)-(1/2)n^2=11/2 -7(n-1)/2 = (18-7n)/2
a(n)=(18-7n+n^2)/2
s(n+1)=2(n+1)-b(n+1)+10
s(n)=2n-b(n)+10
b(n+1)=s(n+1)-s(n)=2-b(n+1)+b(n)
b(n+1)=(1/2)b(n)+1
b(n+1)+x=(1/2)b(n)+1+x=(1/2)[b(n)+2+2x]=(1/2)[b(n)+x]
2+2x=x
x=-2
b(n+1)-2=(1/2)b(n)-1=(1/2)[b(n)-2]
{b(n)-2}是首项为b(1)-2=4,公比为(1/2)的等比数列.
b(n)-2=4*(1/2)^(n-1)=2^(3-n)
b(n)=2 + 2^(3-n)
a(n)-b(n)=(18-7n+n^2)/2 - 2 - 2^(3-n) = (14 - 7n + n^2)/2 - 2^(3-n)
0 < a(n)-b(n) = (14-7n+n^2)/2 - 2^(3-n) < 1/2
0< 14 -7n+n^2-2^(4-n)4时,a(n)-b(n)单调递增,a(n)-b(n)>a(4)-b(4)=1/2.
因此,不存在k,使得0