Difference between revisions of "2023 AMC 12A Problems/Problem 12"
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==Solution 1== | ==Solution 1== | ||
− | + | To solve this problem, we will be using difference of cube, sum of squares and sum of arithmetic sequence formulas | |
− | < | + | <math>2^3-1^3+4^3-3^3+6^3-5^3+...+18^3-17^3</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 22:54, 9 November 2023
Problem
What is the value of
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.