1986 AIME Problem 11

Attempt Problem 11 of the 1986 AIME 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 1986 AIME solutions, or check the answer key.

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11.

The polynomial 1−x+x2−x3+⋯1-x+x^2-x^3+\cdots +x16−x17{}+x^{16}-x^{17} may be written in the form a0+a1y+a2y2+⋯+a16y16+a17y17, \begin{aligned} &a_0+a_1y+a_2y^2+\cdots\\ &\qquad{}+a_{16}y^{16}+a_{17}y^{17}, \end{aligned} where y=x+1y=x+1 and the aia_i’s are constants. Find a2.a_2.
Answer: 816
Concepts:binomial theoremcombinationspolynomial
Small hint:
Substitute x=y−1x=y-1 before expanding
Big hint:
Find only the coefficient of y2y^2 in each power and use the hockey-stick identity
Solution:
The polynomial is ∑k=017(−x)k.\sum_{k=0}^{17}(-x)^k. Since x=y−1,x=y-1, this becomes ∑k=017(1−y)k. \sum_{k=0}^{17}(1-y)^k. For k≥2,k\geq2, the coefficient of y2y^2 in (1−y)k(1-y)^k is (k2).\binom{k}{2}. Hence a2=∑k=217(k2)=(183)=816. a_2=\sum_{k=2}^{17}\binom{k}{2} =\binom{18}{3}=816.
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