1960 AMC 12 Problem 20

Attempt Problem 20 of the 1960 AMC 12 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 1960 AMC 12 solutions, or check the answer key.

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20.

The coefficient of x7x^7 in the expansion of (x22−2x)8 \left(\frac{x^2}{2}-\frac2x\right)^8 is:
5656
−56-56
1414
−14-14
00
Answer: D
Concepts:binomial theoremexponent
Small hint:
In a term using kk copies of −2x,-\frac{2}{x}, determine the exponent of xx
Big hint:
Solve 16−3k=716-3k=7 and then compute that binomial coefficient
Solution:
The term using kk factors of −2x-\frac{2}{x} has exponent 2(8−k)−k=16−3k. 2(8-k)-k=16-3k. To obtain x7,x^7, we need k=3.k=3. Its coefficient is (83)(12)5(−2)3=−14. \binom83\left(\frac12\right)^5(-2)^3=-14.
Thus, the correct answer is D.
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