Difference between revisions of "2014 AMC 12B Problems/Problem 16"
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<cmath>P(2)+P(-2)=12k+2k=\boxed{\textbf{(E)}\ 14k}</cmath> | <cmath>P(2)+P(-2)=12k+2k=\boxed{\textbf{(E)}\ 14k}</cmath> | ||
==Solution 2== | ==Solution 2== | ||
− | If we use Gregory's Triangle, the following happens | + | If we use [[Gregory's Triangle]], the following happens: |
<cmath>P(-1), P(0), P(1)</cmath> | <cmath>P(-1), P(0), P(1)</cmath> | ||
<cmath> 3k , k , 2k </cmath> | <cmath> 3k , k , 2k </cmath> | ||
Line 31: | Line 31: | ||
The above shows us that <math>P(-2)</math> is <math>8k</math> and <math>P(2)</math> is <math>6k</math> so <math>8k+6k=14k</math>. | The above shows us that <math>P(-2)</math> is <math>8k</math> and <math>P(2)</math> is <math>6k</math> so <math>8k+6k=14k</math>. | ||
+ | |||
+ | NOTE (not from author): The link you put for gregory's triangle doesn't work so please explain it in your post or find a resource that does work; there isn't much on google. | ||
== See also == | == See also == | ||
{{AMC12 box|year=2014|ab=B|num-b=15|num-a=17}} | {{AMC12 box|year=2014|ab=B|num-b=15|num-a=17}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 23:34, 1 November 2024
Contents
Problem
Let be a cubic polynomial with , , and . What is ?
Solution
Let . Plugging in for , we find , and plugging in and for , we obtain the following equations: Adding these two equations together, we get If we plug in and in for , we find that Multiplying the third equation by and adding gives us our desired result, so
Solution 2
If we use Gregory's Triangle, the following happens:
Since this is cubic, the common difference is for the linear level so the string of s are infinite in each direction. If we put a on each side of the original , we can solve for and .
The above shows us that is and is so .
NOTE (not from author): The link you put for gregory's triangle doesn't work so please explain it in your post or find a resource that does work; there isn't much on google.
See also
2014 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 15 |
Followed by Problem 17 |
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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