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

Revision as of 23:12, 9 November 2023

What is the value of $2^3 - 1^3 + 4^3 - 3^3 + 6^3 - 5^3 + \dots + 18^3 - 17^3?$

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

To solve this problem, we will be using difference of cube, sum of squares and sum of arithmetic sequence formulas.

$2^3-1^3+4^3-3^3+6^3-5^3+...+18^3-17^3$

$=(2-1)(2^2+1 \cdot 2+1^2)+(4-3)(4^2+4 \cdot 3+3^2)+(6-5)(6^2+6 \cdot 5+5^2)+...+(18-17)(18^2+18 \cdot 17+17^2)$

$=(2^2+1 \cdot 2+1^2)+(4^2+4 \cdot 3+3^2)+(6^2+6 \cdot 5+5^2)+...+(18^2+18 \cdot 17+17^2)$

$=1^2+2^2+3^2+4^2+5^2+6^2...+17^2+18^2+1 \cdot 2+3 \cdot 4+5 \cdot 6+...+17 \cdot 18$

$=\frac{18(18+1)(36+1)}{6}+1 \cdot 2+3 \cdot 4+5 \cdot 6+...+17 \cdot 18$

we could rewrite the second part as $(2n-1)(2n), (1 \leq n \leq 9)$

$(2n-1)(2n)=4n^2-2n$

$\sum{n=1}{9}4n^2=4(\frac{9(9+1)(18+1)}{6})$

$\sum{n=1}{9}-2n=-2(\frac{9(9+1)}{2})$

Hence,

$1 \cdot 2+3 \cdot 4+5 \cdot 6+...+17 \cdot 18 = 4(\frac{9(9+1)(18+1)}{6})-2(\frac{9(9+1)}{2})$

Adding everything up:

$2^3-1^3+4^3-3^3+6^3-5^3+...+18^3-17^3$

$=\frac{18(18+1)(36+1)}{6}+4(\frac{9(9+1)(18+1)}{6})-2(\frac{9(9+1)}{2})$

$=3(19)(37)+6(10)(19)-9(10)$

$=2109+1140-90$

$=\boxed{\textbf{(D) } 3159}$

~lptoggled

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

@@ Line 23: / Line 23: @@
 <math>(2n-1)(2n)=4n^2-2n</math>
-<math>4n^2=4(\frac{9(9+1)(18+1)}{6})</math>
+<math>\sum{n=1}{9}4n^2=4(\frac{9(9+1)(18+1)}{6})</math>
-<math>-2n=-2(\frac{9(9+1)}{2})</math>
+<math>\sum{n=1}{9}-2n=-2(\frac{9(9+1)}{2})</math>
 Hence,