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2023 AMC 12A Problems/Problem 12

Revision as of 02:22, 10 November 2023 by Sqroot (talk | contribs) (→‎Solution 3)

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 $\sum_{n=1}^{9}(2n-1)(2n)$

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

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

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

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

Solution 2

Think about $2^3+4^3+6^3+...+18^3$. Once we factor out $2^3=8$, we get $1^3+2^3+...+9^3$, something which can be easily found using the sum of cubes formula, $(\frac{n(n+1)}{2})^2$. Now think about $1^3+3^3+...+17^3$. This is just the previous sum subtracted from the total sum of 18 cubes. So now we have the two things we need to add. We just need to not screw up the computations: the sum of all the even cubes is just $8\cdot (\frac{90}{2})^2\rightarrow 8\cdot 2025 = 16200$. The sum of all cubes from $1^3$ to $18^3$ is $(\frac{18\cdot 19}{2})^2=29241$. The sum of the odd cubes is then $29241-16200=13041$. Thus we get $16200-13041=\boxed{\textbf{(D) } 3159}$

Solution 3

For any real numbers $x$ and $y$, $x^3-y^3=(x-y)(x^2+xy+y^2)=(x-y)((x-y)^2+3xy)=(x-y)^3+3(x-y)xy$. When $x = y + 1$, we will get $x^3-y^3=(x-y)^3+3(x-y)xy=1 + 3xy$.

Therefore, \[2^3 - 1^3 + 4^3 - 3^3 + 6^3 - 5^3 + \dots + 18^3 - 17^3 = (1 + 3\cdot 1\cdot 2) + (1 + 3\cdot 3\cdot 4) + (1 + 3\cdot 5\cdot 6) + \dots + (1 + 3\cdot 17\cdot 18)\] which is \[\sum_{n=1}^{9}(1 + 3\cdot 2n(2n-1)) = 9 + 3\cdot (1\cdot 2 + 3\cdot 4 + 5\cdot 6 \dots + 17\cdot 18) = 9 + 3 \cdot (2 + 12 + 30 + 56 + 90 + 132 + 182 + 240 + 306) =\boxed{\textbf{(D) } 3159}\]

~sqroot

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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