Problem:
Let Ο=cos72Οβ+iβ
sin72Οβ, where i=β1β. Find the value of the product
k=0β6β(Ο3k+Οk+1)
Solution:
Answer (024):
Note that Ο7=1. If k=0, then the corresponding term in the product is Ο0+Ο0+1=3. For 1β€kβ€3,
(Ο3k+Οk+1)(Ο3(7βk)+Ο7βk+1)β=1+Ο2k+Ο3k+Ο5k+1+Οk+Ο4k+Ο6k+1=2+(1+Ο+Ο2+Ο3+Ο4+Ο5+Ο6)=2β
Thus
k=0β6β(Ο3k+Οk+1)=3k=1β3β(Ο3k+Οk+1)(Ο3(7βk)+Ο7βk+1)=3β
2β
2β
2=24.
OR
Observe that
k=0β6β(Ο3k+Οk+1)β=Ο21k=0β6β(Ο3k+Οk+1)=k=0β6βΟk(Ο3k+Οk+1)β
β=k=0β6β(Οk+Ο2k+Ο4k)=3k=1β6β(Οk+Ο2k+Ο4k)=3(Ο+Ο2+Ο4)3β
(Ο3+Ο5+Ο6)3=3(Ο4+Ο6+Ο7+Ο5+Ο7+Ο8+Ο7+Ο9+Ο10)3=3(Ο4+Ο6+1+Ο5+1+Ο+1+Ο2+Ο3)3=3(1+Ο+Ο2+Ο3+Ο4+Ο5+Ο6+2)3=3β
(0+2)3=24β
OR
Let Ξ±1β,Ξ±2β, and Ξ±3β be the complex roots of x3+x+1, so that
k=0β6β(Ο3k+Οk+1)=k=0β6βj=1β3β(ΟkβΞ±jβ).
Reversing the order in which the products are taken gives
j=1β3βk=0β6β(ΟkβΞ±jβ)=βj=1β3β(Ξ±j7ββ1).
Let Ξ± denote any complex root of x3+x+1. Then
Ξ±7β1β=βΞ±5βΞ±4β1=Ξ±3+Ξ±2+Ξ±2+Ξ±β1=βΞ±β1+2Ξ±2+Ξ±β1=2Ξ±2β2β
Thus the requested product is
j=1β3β(2β2Ξ±j2β)β=β8j=1β3β(1βΞ±jβ)(β1βΞ±jβ)=β8(13+1+1)((β1)3+(β1)+1)=β8β
3β
(β1)=24β
The problems and solutions on this page are the property of the MAA's American Mathematics Competitions