Problem:
Let a1​,a2​,...,an​, be a sequence with the following properties.
(i) a1​=1, and
(ii) a2n​=n⋅an​ for any positive integer n.
What is the value of a2100​?
Answer Choices:
A. 1
B. 299
C. 2100
D. 24950
E. 29999
Solution:
Note that
a21​a22​a23​a24​​=a2​=a2⋅1​=1⋅a1​=20⋅20=20,=a4​=a2⋅2​=2⋅a2​=21⋅20=21,=a8​=a2⋅4​=4⋅a4​=22⋅21=21+2,=a16​=a2⋅8​=8⋅a8​=23⋅21+2=21+2+3​
and, in general, .a2n​=21+2+⋯+(n−1). Because
1+2+3+…+(n−1)=21​n(n−1)
we have a2100​=2(100)(99)/2=24950.
Answer: D​.
The problems on this page are the property of the MAA's American Mathematics Competitions