Difference between revisions of "2023 AMC 12A Problems/Problem 12"
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==Solution 1== | ==Solution 1== | ||
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To solve this problem, we will be using difference of cube, sum of squares and sum of arithmetic sequence formulas | To solve this problem, we will be using difference of cube, sum of squares and sum of arithmetic sequence formulas | ||
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<math>2^3-1^3+4^3-3^3+6^3-5^3+...+18^3-17^3</math> | <math>2^3-1^3+4^3-3^3+6^3-5^3+...+18^3-17^3</math> | ||
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| + | <math>(2-1)(2^2+1 \cdot 2+1^2)+(4-3)(4^2+4 \cdot 3+3^2)+(6-5)(6^2+6 \cdot 5+5^2)+...+(18-17)(18^2+18 \cdot 17+17^2)</math> | ||
==See also== | ==See also== | ||
{{AMC12 box|year=2023|ab=A|num-b=11|num-a=13}} | {{AMC12 box|year=2023|ab=A|num-b=11|num-a=13}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 22:56, 9 November 2023
Problem
What is the value of
![\[2^3 - 1^3 + 4^3 - 3^3 + 6^3 - 5^3 + \dots + 18^3 - 17^3?\]](http://latex.artofproblemsolving.com/9/8/d/98d9501b5431aad136fe4d35c709dc3930eb0858.png)
Solution 1
To solve this problem, we will be using difference of cube, sum of squares and sum of arithmetic sequence formulas
See also
| 2023 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 11 |
Followed by Problem 13 |
| 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 | |
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
