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Difference between revisions of "2019 AMC 12A Problems/Problem 17"

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==Solution 1==
 
==Solution 1==
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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==
 
==Solution 2==
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<math>s_3 - 5s_2 + 8s_1 = 39</math>
 
<math>s_3 - 5s_2 + 8s_1 = 39</math>
  
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<math>39</math> can be written as <math>13s_0</math>.
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<math>39</math> can be written as <math>13s_0</math>, giving
  
 
<math>s_3 = 5s_2 - 8s_1 + 13s_0</math>
 
<math>s_3 = 5s_2 - 8s_1 + 13s_0</math>
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Therefore, <math>a = 5, b = -8</math>, and <math>c = 13</math> by matching coefficients.
 
Therefore, <math>a = 5, b = -8</math>, and <math>c = 13</math> by matching coefficients.
  
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<math>5 - 8 + 13 = \boxed{\textbf{(D)}10}</math>.
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<math>5 - 8 + 13 = \boxed{\textbf{(D) } 10}</math>.
  
 
==Solution 3==
 
==Solution 3==
  
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Let <math>p, q</math>, and <math>r</math> be the roots of the polynomial. By Vieta's, we have
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Let <math>p, q</math>, and <math>r</math> be the roots of the polynomial. By Viète's Formulae, we have
  
 
<math>p+q+r = 5</math>
 
<math>p+q+r = 5</math>
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<math>pqr=13</math>.
 
<math>pqr=13</math>.
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We know <math>s_k = p^k + q^k + r^k</math>. Consider <math>(p+q+r)(s_k) =5s_k</math>.
 
We know <math>s_k = p^k + q^k + r^k</math>. Consider <math>(p+q+r)(s_k) =5s_k</math>.
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<math>13s_{k-2} = p^{k-1}qr + pq^{k-1}r + pqr^{k-1}</math>.
 
<math>13s_{k-2} = p^{k-1}qr + pq^{k-1}r + pqr^{k-1}</math>.
  
−
 
+
We have <cmath>\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) \\
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<math>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)</math>
+
&= 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) \\
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<math> = s_{k+1} + p^{k-1} (pq + pr + qr) + q^{k-1} (pq + pr + qr) + r^{k-1} (pq + pr + qr)</math>
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&= 5s_k + 13s_{k-2} = s_{k+1} + 8s_{k-1}\end{split}</cmath>
−
 
 
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<math> = s_{k+1} + (p^{k-1} + q^{k-1} + r^{k-1})(pq + pr + qr)</math>
 
−
 
 
−
<math> 5s_k + 13s_{k-2} = s_{k+1} + 8s_{k-1}</math>
 
  
 
Rearrange to get
 
Rearrange to get
 
<math>s_{k+1} = 5s_k - 8s_{k-1} + 13s_{k-2}</math>
 
<math>s_{k+1} = 5s_k - 8s_{k-1} + 13s_{k-2}</math>
  
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So, <math>a+ b + c = 5 -8 + 13 = \boxed{\textbf{(D)}10}</math>.
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So, <math>a+ b + c = 5 -8 + 13 = \boxed{\textbf{(D) } 10}</math>.
  
−
- gregwwl
+
-gregwwl
  
 
==Video Solution==
 
==Video Solution==
 
For those who want a video solution: https://www.youtube.com/watch?v=tAS_DbKmtzI
 
For those who want a video solution: https://www.youtube.com/watch?v=tAS_DbKmtzI
−
 
  
 
==See Also==
 
==See Also==

Revision as of 16:53, 18 August 2020

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 Viète'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

Video Solution

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

See Also

2019 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 16 		Followed by
Problem 18
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

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