2021 AMC 12A Problems/Problem 9
- The following problem is from both the 2021 AMC 10A #10 and 2021 AMC 12A #9, so both problems redirect to this page.
Contents
- 1 Problem
- 2 Solution 1
- 3 Solution 2
- 4 Solution 3
- 5 Solution 4 (Engineer's Induction)
- 6 Solution 5 (Difference of Squares)
- 7 Solution 6 (Generalization)
- 8 Solution 7 (Generalization)
- 9 Video Solution
- 10 Video Solution by Aaron He
- 11 Video Solution (Conjugation, Difference of Squares)
- 12 Video Solution by Hawk Math
- 13 Video Solution by OmegaLearn (Factorizations/Telescoping & Meta-solving)
- 14 Video Solution by WhyMath
- 15 Video Solution by TheBeautyofMath
- 16 Video Solution by The Learning Royal
- 17 See also
Problem
Which of the following is equivalent to
Solution 1
By multiplying the entire equation by , all the terms will simplify by difference of squares, and the final answer is .
Additionally, we could also multiply the entire equation (we can let it be equal to ) by . The terms again simplify by difference of squares. This time, we get . Both solutions yield the same answer.
~BakedPotato66
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 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
Solution 5 (Difference of Squares)
We notice that the first term is equal to . If we multiply this by the second term, then we will get , and we can simplify by using difference of squares to obtain . If we multiply this by the third term and simplify using difference of squares again, we get . We can continue down the line until we multiply by the last term, , and get .
~mathboy100
Solution 6 (Generalization)
Let’s generalize to and . Then we get: We see that the first term is , and the next is . We realize that if we multiply the first term by , the result by difference of squares will be : we can proceed to use difference of squares on that. In other words, the equation has the domino effect, and all we need to get started is to multiply the whole equation by (keeping in mind to divide at the end.)
Now we get: Now we can plug in and However, we must not forget to divide by at the end! Dividing, we get the answer of ~KingRavi
Solution 7 (Generalization)
More generally, we have by the difference of squares.
Multiplying the original expression by and then applying repeatedly, we get ~MRENTHUSIASM
Video Solution
-Education the Study of Everything
Video Solution by Aaron He
https://www.youtube.com/watch?v=xTGDKBthWsw&t=9m30s
Video Solution (Conjugation, Difference of Squares)
https://www.youtube.com/watch?v=gXaIyeMF7Qo&list=PLexHyfQ8DMuKqltG3cHT7Di4jhVl6L4YJ&index=9
Video Solution by Hawk Math
https://www.youtube.com/watch?v=P5al76DxyHY
Video Solution by OmegaLearn (Factorizations/Telescoping & Meta-solving)
~ pi_is_3.14
Video Solution by WhyMath
~savannahsolver
Video Solution by TheBeautyofMath
https://youtu.be/s6E4E06XhPU?t=771 (for AMC 10A)
https://youtu.be/cckGBU2x1zg?t=548 (for AMC 12A)
~IceMatrix
Video Solution by The Learning Royal
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.