But you can also try.
an=OA×OBn
Your expression may be unclear. Vector can be divided into point multiplication and cross multiplication. After the cross is still a vector. I guess you want to express point multiplication.
an=OA×OBn
= (p，5) * [n(2/3)^n，0]
= pn(2/3)^n
OA - OCk = (p，5) - (0，k) = (p，5-k)
bk = |OA - OCk|^2 = p^2 + (5 - k)^2
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Berkean
= p^2 + (5 - k)^2 - pn(2/3)^n
If any positive integer n, k, there is always bk-an >；; 1/9 holds, that is, the inequality holds even for the minimum value of the above formula. To this end, find the minimum value of the above formula.
Obviously k = 5
The problem is transformed into finding the maximum value of f (n) = n (2/3) n.
f(n+ 1)/f(n)=(2/3)[(n+ 1)/n]= 2/3+2/(3n)
f(2)/f( 1)> 1
f(3)/f(2) = 1
f(3) = f(2)>f( 1)
After n ≥ 3
f(n+ 1)/f(n)& lt； 1
f(n+ 1)& lt； coedna
So f(2) = f(3) = 8/9 is the maximum value of f(n).
bk - an ≥ p^2 + 0 - p*(8/9)
p^2-8p/9 & gt； 1/9
(p - 1)(p + 1/9)>0
P> 1 and p <-1/9
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bk-an = p^2+(5-k)^2-pn(2/3)^n & lt； 1/9
There is a positive integer n, k, which makes bk-an
This question is a bit weird. I can't understand the questioner's intention, because there will be many nkp that meet the requirements of the topic, and only one set of questions can be given.
According to the previous question, it can be deduced that when k = 5 and n = 2, the value of p can have a range (-1/9, 1).
b5-a2 = p^2-8p/9 & lt； 1/9
(p- 1)(p+ 1/9)& lt； 0
Take p = 0.
b5-a2 = 0 & lt； 1/9 can meet the requirements of the topic.