Question

设｛an}和｛bn}满足a1＝b1＝6,a2＝b2＝4,a3＝b3＝3

Answer 1

1）

a2-a1=4-6=-2

a3-a2=3-4=-1

(a3-a2)-(a2-a1)=-1-(-2)=1

〔an+1－an〕是公差为1的等差数列

an+1-an==-2+(n-1)=n-3

an-a(n-1)=(n-1)-3=n-4

an=a(n-1)+(n-4)

=a(n-2)+(n-5)+(n-4)

=...

=a1+(-2)+(-1)+...+(n-5)+(n-4)

=a1+(n-1)(n-6)/2

=6+(n^2-7n+6)/2

=(n^2-7n+18)/2

b1-2=4

b2-2=2

b3-2=1

数列｛bn－2｝是公比为1/2的等比数列

bn-2=(b1-2)*(1/2)^(n-1)=4*(1/2)^(n-1)=(1/2)^(n-3)

bn=2+(1/2)^(n-3)

2)

Ak－Bk=(k^2-7k+18)/2-2-(1/2)^(k-3)

=(k^2-7k+14)/2-(1/2)^(k-3)