Problem:
The sequence S1β,S2β,S3β,β¦,S10β has the property that every term beginning with the third is the sum of the previous two. That is,
Snβ=Snβ2β+Snβ1β for nβ₯3
Suppose that S9β=110 and S7β=42. What is S4β ?
Answer Choices:
A. 4
B. 6
C. 10
D. 12
E. 16
Solution:
Note that 110=S9β=S7β+S8β=42+S8β, so S8β=110β42=68. Thus 68=S8β=S6β+S7β=S6β+42, so S6β=68β42=26. Similarly, S5β=42β26=16, and S4β=26β16=10β.
The problems on this page are the property of the MAA's American Mathematics Competitions