Difference between revisions of "2023 AMC 12A Problems/Problem 12"

(Solution)
(Solution 1)
Line 6: Line 6:
  
 
==Solution 1==
 
==Solution 1==
 
+
To solve this problem, we will be using difference of cube, sum of squares and sum of arithmetic sequence formulas
<cmath>2^3-1^3+4^3-3^3+6^3-5^3+...+18^3-17^3</cmath>
+
<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 \[2^3 - 1^3 + 4^3 - 3^3 + 6^3 - 5^3 + \dots + 18^3 - 17^3?\]

$\textbf{(A) } 2023 \qquad\textbf{(B) } 2679 \qquad\textbf{(C) } 2941 \qquad\textbf{(D) } 3159 \qquad\textbf{(E) } 3235$

Solution 1

To solve this problem, we will be using difference of cube, sum of squares and sum of arithmetic sequence formulas $2^3-1^3+4^3-3^3+6^3-5^3+...+18^3-17^3$

See also

2023 AMC 12A (ProblemsAnswer KeyResources)
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. AMC logo.png