2019 AMC 12A Problems/Problem 17: Difference between revisions

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Problem

Let $s_k$ denote the sum of the $\textit{k}$ th powers of the roots of the polynomial $x^3-5x^2+8x-13$ . In particular, $s_0=3$ , $s_1=5$ , and $s_2=9$ . Let $a$ , $b$ , and $c$ be real numbers such that $s_{k+1} = a \, s_k + b \, s_{k-1} + c \, s_{k-2}$ for $k = 2$ , $3$ , $....$ What is $a+b+c$ ?

$\textbf{(A)} \; -6 \qquad \textbf{(B)} \; 0 \qquad \textbf{(C)} \; 6 \qquad \textbf{(D)} \; 10 \qquad \textbf{(E)} \; 26$

Solution 1

Applying Newton's Sums (see this link), we have $s_{k+1}+(-5)s_k+(8)s_{k-1}+(-13)s_{k-2}=0,$ so $s_{k+1}=5s_k-8s_{k-1}+13s_{k-2},$ we get the answer as $5+(-8)+13=10$ .

Solution 2

Let $p, q$ , and $r$ be the roots of the polynomial. Then,

$p^3 - 5p^2 + 8p - 13 = 0$

$q^3 - 5q^2 + 8q - 13 = 0$

$r^3 - 5r^2 + 8r - 13 = 0$

Adding these three equations, we get

$(p^3 + q^3 + r^3) - 5(p^2 + q^2 + r^2) + 8(p + q + r) - 39 = 0$

$s_3 - 5s_2 + 8s_1 = 39$

$39$ can be written as $13s_0$ , giving

$s_3 = 5s_2 - 8s_1 + 13s_0$

We are given that $s_{k+1} = a \, s_k + b \, s_{k-1} + c \, s_{k-2}$ is satisfied for $k = 2$ , $3$ , $....$ , meaning it must be satisfied when $k = 2$ , giving us $s_3 = a \, s_2 + b \, s_1 + c \, s_0$ .

Therefore, $a = 5, b = -8$ , and $c = 13$ by matching coefficients.

$5 - 8 + 13 = \boxed{\textbf{(D) } 10}$ .

Solution 3

Let $p, q$ , and $r$ be the roots of the polynomial. By Vieta's Formulae, we have

$p+q+r = 5$

$pq+qr+rp = 8$

$pqr=13$ .

We know $s_k = p^k + q^k + r^k$ . Consider $(p+q+r)(s_k) =5s_k$ .

$5s_k = [p^{k+1} + q^{k+1} + r^{k+1}] + p^k q + p^k r + pq^k + q^k r + pr^k + qr^k$

Using $pqr = 13$ and $s_{k-2} = p^{k-2} + q^{k-2} + r^{k-2}$ , we see $13s_{k-2} = p^{k-1}qr + pq^{k-1}r + pqr^{k-1}$ .

We have $\begin{split} 5s_k + 13s_{k-2} &= s_{k+1} + (p^k q + p^k r + p^{k-1}qr) + (pq^k + pq^{k-1}r + q^k r) + (pqr^{k-1} + pr^k + qr^k) \\ &= s_{k+1} + p^{k-1} (pq + pr + qr) + q^{k-1} (pq + pr + qr) + r^{k-1} (pq + pr + qr) \\ &= s_{k+1} + (p^{k-1} + q^{k-1} + r^{k-1})(pq + pr + qr) \\ &= 5s_k + 13s_{k-2} = s_{k+1} + 8s_{k-1}\end{split}$

Rearrange to get $s_{k+1} = 5s_k - 8s_{k-1} + 13s_{k-2}$

So, $a+ b + c = 5 -8 + 13 = \boxed{\textbf{(D) } 10}$ .

Video Solution

For those who want a video solution: https://www.youtube.com/watch?v=tAS_DbKmtzI

@@ Line 6: / Line 6: @@
 ==Solution 1==
-Applying Newton's Sums (see [https://artofproblemsolving.com/wiki/index.php/Newton%27s_Sums this link]), we get the answer as <math>10</math>.
+Applying Newton's Sums (see [https://artofproblemsolving.com/wiki/index.php/Newton%27s_Sums this link]), we have<cmath>s_{k+1}+(-5)s_k+(8)s_{k-1}+(-13)s_{k-2}=0,</cmath>so<cmath>s_{k+1}=5s_k-8s_{k-1}+13s_{k-2},</cmath>we get the answer as <math>5+(-8)+13=10</math>.
 ==Solution 2==

2019 AMC 12A (Problems • Answer Key • Resources)
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