Problem:
Let f(n)=(2β1+i3ββ)n+(2β1βi3ββ)n, where i=β1β. What is f(2022) ?
Answer Choices:
A. β2
B. β1
C. 0
D. 3β
E. 2
Solution:
Note that
2β1+i3ββ=cos32Οβ+isin32Οβ=e32Οiβ
and
2β1βi3ββ=cos3β2Οβ+isin3β2Οβ=e3β2Οiβ
By de Moivre's Formula,
f(n)=e32nΟiβ+e3β2nΟiβ
Therefore
f(2022)=e1348Οi+eβ1348Οi
Because e2Οi=1, the given expression equals 1+1=(E)2β.
The problems on this page are the property of the MAA's American Mathematics Competitions