Difference between revisions of "2021 AMC 12A Problems/Problem 9"
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==Problem== | ==Problem== | ||
Which of the following is equivalent to | Which of the following is equivalent to |
Revision as of 18:39, 12 February 2021
- The following problem is from both the 2021 AMC 10A #10 and 2021 AMC 12A #9, so both problems redirect to this page.
Contents
Problem
Which of the following is equivalent to
Solution 1
All you need to do is multiply the entire equation by . Then all the terms will easily simplify by difference of squares and you will get or as your final answer. Notice you don't need to worry about because that's equal to .
-Lemonie
Solution 2
If you weren't able to come up with the insight, then you could just notice that the answer is divisible by , and . We can then use Fermat's Little Theorem for on the answer choices to determine which of the answer choices are divisible by both and . This is .
-MEWTO
Solution 3
After expanding the first few terms, the result after each term appears to be where n is the number of terms expanded. We can prove this using mathematical induction. The base step is trivial. When expanding another term, all of the previous terms multiplied by would give , and all the previous terms multiplied by would give . Their sum is equal to , so the proof is complete. Since is equal to , the answer is .
-SmileKat32
Solution 4 (Engineer's Induction)
We can compute some of the first few partial products, and notice that . As we don't have to prove this, we get the product is , and smugly click . ~rocketsri
Video Solution by Hawk Math
https://www.youtube.com/watch?v=P5al76DxyHY
Video Solution by OmegaLearn(Factorizations/Telescoping& Meta-solving)
~ pi_is_3.14
See also
2021 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 9 |
Followed by Problem 11 | |
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 10 Problems and Solutions |
2021 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 8 |
Followed by Problem 10 |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.