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
a21a22a23a24=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)=21n(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