1993 AIME Problems/Problem 5

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Problem

Let $P_0(x) = x^3 + 313x^2 - 77x - 8\,$ . For integers $n \ge 1\,$ , define $P_n(x) = P_{n - 1}(x - n)\,$ . What is the coefficient of $x\,$ in $P_{20}(x)\,$ ?

Solution

Notice that $\begin{align*}P_{20}(x) &= P_{19}(x - 20)\\ &= P_{18}((x - 20) - 19)\\ &= P_{17}(((x - 20) - 19) - 18)\\ &= \cdots\\ &= P_0(x - (20 + 19 + 18 + \ldots + 2 + 1)).\end{align*}$

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 $630 \cdot 210 - 626 \cdot 210 - 77 = \boxed{763}$ .

1993 AIME (Problems • Resources)
Preceded by Problem 4	Followed by Problem 6
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

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1993 AIME Problems/Problem 5

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Problem

Solution

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