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
yn​yn+1​​=1−yn​+101​
it follows that
109​≤yn​yn+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