2019 AMC 12A Problems/Problem 17

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

-gregwwl

Solution 4

Let $r,s,t$ be the roots of $x^3-5x^2+8x-13$ . Then:

$r^3=5r^2-8r+13$ \\ $s^3=5s^2-8s+13$ \\ $t^3=5t^2-8t+13$

If we multiply both sides of the equation by $r^k$ , where $k$ is a positive integer, then that won't change the coefficients, but just the degree of the new polynomial and the other term's exponents. We can try multiplying to find $r^4+s^4+t^4$ , but that is just to check. So then with the above information about $r^3,s^3,t^3$ , we see that:

$r^k=5r^{k-1}-8r^{k-2}-13r^{k-3}$ , $s^k=5s^{k-1}-8s^{k-2}-13s^{k-3}$ , $t^k=5t^{k-1}-8t^{k-2}-13t^{k-3}$

$s_k=r^k+s^k+t^k$

Then: $s_k=5s_{k-1}-8s_{k-2}+13s_{k-3}$

This means that $s_{k+1}=5s_{k}-8s_{k-1}+13s_{k-2}$ , as expected. So we have $a=5, b=-8, c=13$ . So our answer is $5-8+13=\boxed{\textbf{(D) } 10}$

-IzhanAli

Solution 5

Let the roots be $r$ , $s$ , and $t$ . We know $r^2+s^2+t^2 = (r+s+t)(r+s+t) - 2(rs+st+tr)$ .Continuing, we have:

$r^3+s^3+t^3 = (r^2+s^2+t^2)(r+s+t) - (rs+st+tr)(r+s+t)+3rst$

$r^4+s^4+t^4 = (r^3+s^3+t^3)(r+s+t) - (rs+st+tr)(r^2+s^2+t^2)-(rst)(r+s+t)$

$r^5+s^5+t^5 = (r^4+s^4+t^4)(r+s+t) - (rs+st+tr)(r^3+s^3+t^3) - (rst)(r^2+s^2+t^2)$

Clearly, the answer is $5-8+13 = \boxed{\textbf{(D)} 10}$

-skibbysiggy

Video Solution

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

2019 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 16	Followed by Problem 18
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