Problem:
A sequence (a1,b1),(a2,b2),(a3,b3),… of points in the coordinate plane satisfies
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(an+1,bn+1)=(3an−bn,3bn+an) for n=1,2,3,…
Suppose that (a100,b100)=(2,4). What is a1+b1?
Answer Choices:
A. −2971
B. −2991
C. 0
D. 2981
E. 2961
Solution:
Let zn=an+bni. Then
zn+1=(3an−bn)+(3bn+an)i=(an+bni)(3+i)=zn(3+i)=z1(3+i)n
Noting that 3+i=2(cos6π+isin6π) and applying DeMoivre's formula gives
2+4i=z100=z1(2(cos(6π)+isin(6π)))99=z1⋅299(cos(699π)+isin(699π))=(a1+b1i)⋅299⋅i=−299b1+299a1i
So 2=−299b1,4=299a1, and
a1+b1=2994−2992=2981
Note that
(an+2,bn+2)=(3(3an−bn)−(3bn+an)3(3bn+an)+(3an−bn))=(−23bn+2an,23an+2bn)(an+3,bn+3)=(3(−23bn+2an)−(23an+2bn)3(23an+2bn)+(−23bn+2an))=8(−bn,an)
and (an+6,bn+6)=8(−bn+3,an+3)=−64(an,bn). Because 97=1+16⋅6, we have
(a97,b97)=(−64)16(a1,b1)=296(a1,b1)
and
(2,4)=(a100,b100)=23(−b97,a97)=299(−b1,a1)
The conclusion follows as in the first solution.
The problems on this page are the property of the MAA's American Mathematics Competitions