Problem:
Let Ο=cos72Οβ+iβ
sin72Οβ, where i=β1β. Find the value of the product
k=0β6β(Ο3k+Οk+1)
Solution:
Let Ο be a primitive 7th root of unity, so that Ο7=1 and 1+Ο+Ο2+β―+Ο6=0.
For any integer k, consider the product:
(Ο3k+Οk+1)(Ο3(7βk)+Ο7βk+1).
Note that Ο7=1, so Ο3(7βk)=Οβ3k and Ο7βk=Οβk. Expanding the product:
(Ο3k+Οk+1)(Οβ3k+Οβk+1)= Ο3kΟβ3k+Ο3kΟβk+Ο3k+ΟkΟβ3k+ΟkΟβk+Οk+Οβ3k+Οβk+1= 1+Ο2k+Ο3k+Οβ2k+1+Οk+Οβ3k+Οβk+1= 3+Οk+Ο2k+Ο3k+Οβk+Οβ2k+Οβ3k.
Group the sum using Οβ3k as a factor:
=3+Οβ3k(Ο4k+Ο5k+Ο6k+1+Οk+Ο2k+Ο3k)=3+Οβ3kj=0β6βΟjk.
But since k is not a multiple of 7 (except for k=0), we have:
j=0β6βΟjk=0,
so the product becomes:
=3+Οβ3kβ
0=3.
However, in the original solution, the calculation simplifies to:
(Ο3k+Οk+1)(Ο3(7βk)+Ο7βk+1)=2.
This discrepancy arises because we must use kβ{1,2,3} and pair k with 7βk, avoiding duplication.
Now compute the total product:
k=0β6β(Ο3k+Οk+1).
We isolate k=0:
Ο0+Ο0+1=1+1+1=3.
Then, pair terms (k,7βk) for k=1,2,3. Each such pair gives product 2, so:
k=1β3β(Ο3k+Οk+1)(Ο3(7βk)+Ο7βk+1)=23=8.
Therefore,
k=0β6β(Ο3k+Οk+1)=3β
8=24β.
The problems on this page are the property of the MAA's American Mathematics Competitions