# 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 $$3 \cdot 231^2 + 213 \cdot 2 \cdot -231 - 67$$ $$231(231 \cdot 3 - 426) - 67$$ $$231(693-426)-67$$ $$231(267)-67$$ $$\boxed{61610}$$