设数列{an}和{bn}满足a1=b1=6,a2=b2=4,a3=b3=3且数列{a(n+1)-an}是等差数列,数列{

(1){a(n+1)-an}是等差数列
设Cn=a(n+1)-an
则C1=a2-a1=4-6=-2
C2=a3-a2=3-4=-1
d=C2-C1=1
Cn=C1+(n-1)d=n-3
Sn=(C1+Cn)*n/2=(n-5)n/2=a2-a1+a3-a2+...+a(n+1)-an=a(n+1)-a1=a(n+1)-6
a(n+1)-6=(n-5)n/2
a(n+1)=(n-5)n/2+6
an=(n-6)(n-1)/2+6=1/2n^2-7/2n+9
设Dn=bn-2是等比数列
则D1=4
D2=b2-2=2
q=D2/D1=1/2
Dn=D1*q^(n-1)=2*1/2^(n-1)=1/2^(n-2)=bn-2
bn=1/2^(n-2)+2
(2)k=1,2,3时,bk-ak=0
k=4时,
bk=9/4
ak=3
bk-ak=-3/4
当k>4时,因bk是减函数,所以bk3
所以bk-ak