Problem:
Let (anβ) and (bnβ) be the sequences of real numbers such that
(2+i)n=anβ+bnβi
for all integers nβ₯0, where i=β1β. What is
n=0βββ7nanβbnββ?
Answer Choices:
A. 83β
B. 167β
C. 21β
D. 169β
E. 74β
Solution:
Square the given equality to yield
(3+4i)n=(anβ+bnβi)2= (an2ββbn2β)+2anβbnβi
so anβbnβ= 21βIm((3+4i)n) and
βnβ₯0β7nanβbnββ= 21βIm(βnβ₯0β7n(3+4i)nβ)= 21βImβββββ1β73+4iβ1ββ ββββ= (B)167ββ
OR
Note that (2+i)= 5ββ
(5β2β+5β1βi). Let ΞΈ=arctan(1/2), then, we know that
(2+i)= 5ββ
(cosΞΈ+isinΞΈ)
so
(2+i)n= (cos(nΞΈ)+isin(nΞΈ))(5β)n .
Therefore,
βn=0ββ7nanβbnββ= βn=0ββ7ncos(nΞΈ)sin(nΞΈ)(5)nβ= 21ββn=0ββ(75β)nsin(2nΞΈ)
=$ 21βIm(βn=0ββ(75β)ne2iΞΈn)
Aha! βn=0ββ(75β)ne2iΞΈn is a geometric sequence that evaluates to 1β75βe2ΞΈi1β ! Now we can quickly see that
sin(2ΞΈ)= 2β
sinΞΈβ
cosΞΈ= 2β
5β2ββ
5β1β= \dfrac{4}
cos(2ΞΈ)=cos2ΞΈβsin2ΞΈ= 54ββ51β= \dfrac{3}
Therefore,
1β75βe2ΞΈi1β=1β75β(53β+54βi)1β= 87β+87βi
The imaginary part is 87β, so our answer is 21ββ
87β= (B)167ββ.
The problems on this page are the property of the MAA's American Mathematics Competitions