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

(→‎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
  
−
<math>2^3-1^3+4^3-3^3+6^3-5^3+...+18^3-17^3</math>
+
<cmath>2^3-1^3+4^3-3^3+6^3-5^3+...+18^3-17^3</cmath>
  
 
<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>
 
<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>
Line 23: Line 23:
 
<math>(2n-1)(2n)=4n^2-2n</math>
 
<math>(2n-1)(2n)=4n^2-2n</math>
  
−
<math>4n^2=4(\frac{9(9+1)(18+1)}{6})</math> <math>-2n=-2(\frac{9(9+1)}{2})</math>
+
<math>4n^2=4(\frac{9(9+1)(18+1)}{6})</math>
 +
 
 +
<math>-2n=-2(\frac{9(9+1)}{2})</math>
  
 
Hence,
 
Hence,
  
−
<math>1 \cdot 2+3 \cdot 4+5 \cdot 6+...+17 \cdot 18</math>
+
<math>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})</math>
−
 
 
−
<math>=4(\frac{9(9+1)(18+1)}{6})-2(\frac{9(9+1)}{2})</math>
 
  
 
Adding everything up:
 
Adding everything up:
Line 35: Line 35:
 
<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>
 
<math>=\frac{18(18+1)(36+1)}{6}+4(\frac{9(9+1)(18+1)}{6})-2(\frac{9(9+1)}{2})</math>
 
<math>=\frac{18(18+1)(36+1)}{6}+4(\frac{9(9+1)(18+1)}{6})-2(\frac{9(9+1)}{2})</math>
 +
<math>=3(19)(37)+6(10)(19)-9(10)</math>
 +
<math>=2109+1140-90</math>
 +
$=\boxed{\textbf{(D) } 3159}
  
 
==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 23:10, 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

\[2^3-1^3+4^3-3^3+6^3-5^3+...+18^3-17^3\]

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

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

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

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

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

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

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

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

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}

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

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