Problem:
Given that
x1=211x2=375x3=420x4=523, and xn=xn−1−xn−2+xn−3−xn−4 when n≥5
find the value of x531+x753+x975.
Solution:
For n>5,
xn=xn−1−xn−2+xn−3−xn−4=(xn−2−xn−3+xn−4−xn−5)−xn−2+xn−3−xn−4=−xn−5
It follows that the sequence repeats in a cycle ten terms long. Hence
x531+x753+x975=x1+x3+x5=x1+x3+x4−x3+x2−x1=x4+x2=523+375=898.
OR
Using the theory of difference equations, a characteristic equation for the sequence is x4=x3−x2+x−1 or x4−x3+x2−x+1=0. Since x5+1=(x+1)(x4−x3+x2−x+1), we can conclude xn+xn−5=0 and proceed as above.
The problems on this page are the property of the MAA's American Mathematics Competitions