Problem:
Define a sequence recursively by x0=5 and
xn+1=xn+6xn2+5xn+4
for all nonnegative integers n. Let m be the least positive integer such that
xm≤4+2201
In which of the following intervals does m lie?
Answer Choices:
A. [9,26]
B. [27,80]
C. [81,242]
D. [243,728]
E. [729,∞]
Solution:
First note that it suffices to study yn=xn−4 and find the least positive integer m such that ym≤2201. Now y0=1 and
yn+1=yn+10yn(yn+9)
Observe that (yn) is a strictly decreasing sequence of positive numbers. Because
ynyn+1=1−yn+101
it follows that
109≤ynyn+1≤1110
and because y0=1,
(109)k≤yk≤(1110)k
for all integers k≥2.
Now note that
(21)41<109
because this is equivalent to 0.5<(0.9)4=(0.81)2. Therefore
(21)4m<ym≤2201
so m>80. Now note that
(1011)10=(1+101)10>1+10⋅101=2
so
1110<(21)101
Thus
2201<ym−1<(1110)m−1<(21)10m−1
so m<201. Thus m lies in the range (C). (Numerical calculations will show that m=133.)
The problems on this page are the property of the MAA's American Mathematics Competitions