2000 Pan African MO Problems/Problem 2

Problem

Define the polynomials $P_0, P_1, P_2 \cdots$ by: $P_0(x)=x^3+213x^2-67x-2000$ $P_n(x)=P_{n-1}(x-n), n \in N$ Find the coefficient of $x$ in $P_{21}(x)$ .

Solution

Note that $P_1(x) = P_0(x-1)$ , $P_2(x) = P_1(x-2) = P_0(x-3)$ , and $P_3(x) = P_2(x-3) = P_1(x-5) = P_0(x-6)$ . Thus, we can suspect that $P_n(x) = P_0(x - \tfrac{n(n+1)}{2})$ and use induction to prove it.

The base case works because $P_1(x) = P_0(x - \tfrac{1 \cdot 2}{2}) = P_0(x-1)$ . For the inductive step, assume that $P_n(x) = P_0(x - \tfrac{n(n+1)}{2})$ . Thus, $\begin{align*} P_n(x - (n+1)) &= P_0(x - (n+1) - \tfrac{n(n+1)}{2}) \\ P_{n+1}(x) &= P_0(x - \tfrac{(n+1)(n+2)}{2}) \end{align*}$

Therefore, $P_n(x) = P_0(x - \tfrac{n(n+1)}{2})$ , so $\begin{align*} P_{21}(x) &= P_0(x - 21 \cdot 11) \\ &= P_0(x-231) \\ &= (x-231)^3 + 213(x-231)^2 - 67(x-231) - 2000 \end{align*}$

To find the coefficient of $x$ , we can use the Binomial Theorem. The coefficient of $x$ is $3 \cdot 231^2 + 213 \cdot 2 \cdot -231 - 67$ $231(231 \cdot 3 - 426) - 67$ $231(693-426)-67$ $231(267)-67$ $\boxed{61610}$

2000 Pan African MO (Problems)
Preceded by Problem 1	1 • 2 • 3 • 4 • 5 • 6	Followed by Problem 3
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2000 Pan African MO Problems/Problem 2

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