Question

数列{an}与{bn},an=(1/2)n^2-7n+18,bn=2+(1/2)^(n-3),是否存在K属于正整数,使a

Answer 1

an-bn=(1/2)n^2-7n+18-2-(1/2)^(n-3)

=(1/2)n^2-(1/2)^(n-3)-7n+16

n=1或n=2,n=3时

an-bn>1

n>3时,1>(1/2)^(n-3)>0

(1/2)n^2-7n+16 >an-bn>(1/2)n^2-1-7n+16

(1/2)(n-7)^2-17/2>an-bn>(1/2)(n-7)^2-19/2

n=11

(-1/2)>an-bn>(-3/2)

n=12

4>an-bn>3

因此不存在整数k

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