Problem:
Let sk denote the sum of the k th powers of the roots of the polynomial x3−5x2+8x−13. In particular, s0=3,s1=5, and s2=9. Let a,b, and c be real numbers such that sk+1= ask+bsk−1+csk−2 for k=2,3,… What is a+b+c?
Answer Choices:
A. −6
B. 0
C. 6
D. 10
E. 26
Solution:
Applying Newton's Sums, we have
sk+1+(−5)sk+(8)sk−1+(−13)sk−2=0
so
s_{k+1}=5 s_{k}-8 s_{k-1}+13 s_
we get the answer as 5+(−8)+13=(D)10 .
OR
Let p,q, and r be the roots of the polynomial. Then
p3−5p2+8p−13=0
q3−5q2+8q−13=0
r3−5r2+8r−13=0
Adding these three equations, we get
(p3+q3+r3)−5(p2+q2+r2)+8(p+q+r)−39=0
s3−5s2+8s1=39
39 can be written as 13s0, giving
s_{3}=5 s_{2}-8 s_{1}+13 s_
We are given that sk+1=ask+bsk−1+csk−2 is satisfied for k=2,3,…. , meaning it must be satisfied when k=2, giving us s3=as2+bs1+cs0.
Therefore, a=5,b=−8, and c=13 by matching coefficients.
5−8+13= (D)10 .
The problems on this page are the property of the MAA's American Mathematics Competitions