2023 AMC 12A Problem 12 — Solution | LIVE by Po-Shen Loh

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2023 AMC 12A Problem 12

Below is the professionally curated solution for Problem 12 of the 2023 AMC 12A, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2023 AMC 12A solutions, or check the answer key.

All of the real AMC 8, AMC 10, AMC 12, and AIME problems in our complete solution collection are used with official legal permission of the Mathematical Association of America (MAA).

Concepts:sum and difference of cubes sum of first n squares

Difficulty rating: 1630

12.

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

$2023$

$2679$

$2941$

$3159$

$3235$

Solution:

Group into pairs $(2k)^3-(2k-1)^3$ for $k=1,\ldots,9.$ Expanding, $(2k)^3-(2k-1)^3=12k^2-6k+1.$

Summing for $k=1$ to $9,$ with $\sum k^2=285$ and $\sum k=45,$ gives $12\cdot 285-6\cdot 45+9=3420-270+9=3159.$

Thus, the correct answer is D.

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