2023 AMC 12A Problem 12

Attempt Problem 12 of the 2023 AMC 12A below, then check your answer against the professionally curated solution 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 problems are used with official legal permission of the Mathematical Association of America (MAA).

12.

What is the value of 23−13+43−33+63−53+⋯+183−173? \begin{gathered} 2^3-1^3+4^3-3^3+6^3-5^3\\ {}+\cdots+18^3-17^3? \end{gathered}
20232023
26792679
29412941
31593159
32353235
Answer: D
Concepts:sum and difference of cubessum of first n squares
Small hint:
Group the terms into pairs (2k)3−(2k−1)3(2k)^3-(2k-1)^3 for k=1k=1 to 99
Big hint:
Each pair equals 12k2−6k+1;12k^2-6k+1; sum using ∑k2\sum k^2 and ∑k\sum k
Solution:
Group into pairs (2k)3−(2k−1)3(2k)^3-(2k-1)^3 for k=1,…,9.k=1,\ldots,9. Expanding, (2k)3−(2k−1)3=12k2−6k+1. \begin{gathered} (2k)^3-(2k-1)^3\\ {}=12k^2-6k+1. \end{gathered}
Summing for k=1k=1 to 9,9, with ∑k2=285\sum k^2=285 and ∑k=45,\sum k=45, gives 12⋅285−6⋅45+9=3420−270+9=3159. \begin{gathered} 12\cdot 285-6\cdot 45+9\\ {}=3420-270+9\\ {}=3159. \end{gathered}
Thus, the correct answer is D.
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