AIME 1993 · 第 5 题

Solution 1

Notice that

$$\begin{aligned}P_{20}(x) &= P_{19}(x - 20)\\ &= P_{18}((x - 20) - 19)\\ &= P_{17}(((x - 20) - 19) - 18)\\ &= \cdots\\ &= P_0(x - (20 + 19 + 18 + \ldots + 2 + 1)).\end{aligned}$$ Using the formula for the sum of the first $n$ numbers, $1 + 2 + \cdots + 20 = \frac{20(20+1)}{2} = 210$. Therefore,

$$P_{20}(x) = P_0(x - 210).$$ Substituting $x - 210$ into the function definition, we get $P_0(x-210) = (x - 210)^3 + 313(x - 210)^2 - 77(x - 210) - 8$. We only need the coefficients of the linear terms, which we can find by the binomial theorem.

• $(x-210)^3$ will have a linear term of ${3\choose1}210^2x = 630 \cdot 210x$.
• $313(x-210)^2$ will have a linear term of $-313 \cdot {2\choose1}210x = -626 \cdot 210x$.
• $-77(x-210)$ will have a linear term of $-77x$.

Adding up the coefficients, we get $630 \cdot 210 - 626 \cdot 210 - 77 = \boxed{763}$.

Solution 2 (Overkill)

Denote the coefficient of term $x^m$ of polynomial $P_n(x)$ as $c_{m, n}$. We can see that $c_{2, 0} = 313$, $c_{1, 0} = -77$, and $c_{0, 0} = -8$. Note that $(x-n)^3 = x^3-3nx^2+3n^2x-n^3$ and $(x-n)^2 = x^2-2nx+n^2$ When substituting $x-1$ for $x$ in $P_0(x)$, we get that $P_1(x) = P_0(x-1) = (x^3-3x^2+3x-1) + 313(x^2-2x+1) -77(x-1) -8$. Observe that

$$c_{2,1} = c_{2,0} - 3(1) = 313 - 3 = 310$$ It is evident that

$$c_{2,n} = c_{2,n-1} - 3n$$ Define the generating function $C_2(X)$ as

$$C_2(X) := \sum_{n \geq 0} c_{2,n}X^n$$ By multiplying both sides of the previous recurrence relation and taking the sum of the terms $\forall n\geq 1$, we get that

$$\implies \sum_{n\geq1} c_{2,n}X^n = \sum_{n\geq1}c_{2,n-1}X^n + 3\sum_{n\geq1}nX^n$$ $$\implies \sum_{n\geq0} c_{2,n}X^n - c_{2,0} = X\sum_{n\geq0}c_{2,n}X^n + 3\sum_{n\geq1}nX^n$$ $$\implies C_2(X) - 313 = XC_2(X) + 3\cdot \frac{X}{(1-X)^2}^*$$ $$\implies C_2(X) = \frac{313}{1-X} - \frac{3X}{(1-X)^3}$$ $$= \frac{313X^2 - 629X + 313}{(1-X)^3} = (313X^2 - 629X + 313) \cdot \sum_{n\geq0}{n+2 \choose 2}X^n\ ^\dagger$$ $$\implies c_{2,n} = 313{n+2 \choose 2} - 629{n+1 \choose 2} + 313{n \choose 2}^\ddagger = \frac{-3n^2 - 3n + 626}{2}$$ We can perform a similar analysis on $c_{1,n}$ to get the recurrence relation

$$c_{1,n} = c_{1, n-1} + c_{2,n-1}\cdot(-2n) + 3n^2$$ $$= c_{1, n-1} +3n^3 - 626n$$ Define the generating function $C_1(X)$ as

$$C_1(X) := \sum_{n\geq0}c_{1,n}X^n$$ We can then perform this process again on our new recurrence relation:

$$\implies \sum_{n\geq1} c_{1,n}X^n = \sum_{n\geq1}c_{1,n-1}X^n + 3\sum_{n\geq1}n^3X^n - 626\sum_{n\geq1}nX^n$$ $$\implies \sum_{n\geq0} c_{1,n}X^n - c_{1,0} = X\sum_{n\geq0}c_{1,n}X^n + 3\sum_{n\geq1}n^3X^n - 626\sum_{n\geq1}nX^n$$ $$\implies C_1(X) + 77 = XC_1(X) + 3\cdot \frac{X(X^2+4X+1)}{(1-X)^4} - 626\cdot \frac{X}{(1-X)^2}$$ $$\implies C_1(X) = \frac{-77(1-X)^4 + 3X(X^2+4X+1) - 626X(1-X)^2}{(1-X)^5}$$ $$= \frac{-77X^4-315X^3+802X^2-315X-77}{(1-X)^5}$$ $$=(-77X^4-315X^3+802X^2-315X-77) \cdot \sum_{n\geq0}{n+5 \choose 5}X^n$$ $$\implies c_{1,n} = -77{n+4 \choose 4} - 315{n+3 \choose 4} + 802{n+2 \choose 4} - 315{n+1 \choose 4} -77{n \choose 4}$$ $$= \frac{3n^4+6n^3-1249n^2-1252n-308}{4}$$ Finally, we can plug $n=20$ into our new explicit formula to get

$$c_{1,20} = \boxed{763}$$ $^*$This can be calculated by differentiating the generating function of the sequence 1, 1, 1, ... $(\sum_{n\geq0}X^n)$ and multiplying by $X$.

$^\dagger$This form can be found by applying Newton's generalized binomial theorem.

$^\ddagger$This formula can be found by convolving the polynomial with the series.

- MathCactus0_0 (don't try this on a test unless you can't think of anything else!!!)