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