Difference between revisions of "2022 AMC 10B Problems/Problem 15"
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Since the sum of the first <math>n</math> odd numbers is <math>n^2</math>, <math>S_{20} = 20^2 = \boxed{\textbf{(D) } 400}</math>. | Since the sum of the first <math>n</math> odd numbers is <math>n^2</math>, <math>S_{20} = 20^2 = \boxed{\textbf{(D) } 400}</math>. | ||
− | ==Solution | + | ==Solution 2 (Quick Insight)== |
Recall that the sum of the first <math>n</math> odd numbers is <math>n^2</math>. | Recall that the sum of the first <math>n</math> odd numbers is <math>n^2</math>. | ||
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~numerophile | ~numerophile | ||
− | ==Video Solution (🚀 Solved in | + | ==Video Solution (🚀 Solved in 5 min 🚀)== |
https://youtu.be/7ztNpblm2TY | https://youtu.be/7ztNpblm2TY | ||
~Education, the Study of Everything | ~Education, the Study of Everything | ||
+ | |||
==Video Solution By SpreadTheMathLove== | ==Video Solution By SpreadTheMathLove== | ||
https://www.youtube.com/watch?v=zHJJyMlH9DA | https://www.youtube.com/watch?v=zHJJyMlH9DA |
Latest revision as of 14:02, 10 November 2024
Contents
[hide]Problem
Let be the sum of the first terms of an arithmetic sequence that has a common difference of . The quotient does not depend on . What is ?
Solution 1
Let's say that our sequence is Then, since the value of n doesn't matter in the quotient , we can say that Simplifying, we get , from which Solving for , we get that .
Since the sum of the first odd numbers is , .
Solution 2 (Quick Insight)
Recall that the sum of the first odd numbers is .
Since , we have .
~numerophile
Video Solution (🚀 Solved in 5 min 🚀)
~Education, the Study of Everything
Video Solution By SpreadTheMathLove
https://www.youtube.com/watch?v=zHJJyMlH9DA
Video Solution by Interstigation
Video Solution by paixiao
https://www.youtube.com/watch?v=4bzuoKi2Tes
Video Solution by TheBeautyofMath
https://youtu.be/Mi2AxPhnRno?t=1299
See Also
2022 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Problem 16 | |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.