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.
问题描述：
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
答
a(n+1)-a(n)=a+(n-1)da=a(2)-a(1)=4-6=-2a+d=a(3)-a(2)=3-4=-1d=-1-a=1a(n+1)-a(n)=-2+n-1=n-3a(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/...