Let k car be the upper limit, and the relationship should be satisfied. According to the meaning of the question, the recursive formula is obtained;
a 1=30
a(n+ 1)=an*( 1-6%) + k
Judging from the meaning of the problem, it is necessary to make any n have one
The following derivation leads to this formula.
a 1=30
A (n+1) = an * (1-6%)+k-formula1.
Structural geometric series, setting
a(n+ 1)+ t = 0.96*(an + t)
Reference equation 1
t=- 250k
That is, [a (n+1)-250k]/[an-250k] = 0.96.
For geometric series.
therefore
An = (a1-250k) * (0.96) (n-1)
=(30 -250k)*(0.96)^(n- 1)+250k
One; one
Obviously, the bigger n is, the smaller an is.
When n=2
Therefore (30-250k) * 0.96+250k < = 60.
solve
k & lt=3. 12
Taking k=3 is what you want.