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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)\\ &= \...


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}.

See also

1993 AIME (ProblemsResources)
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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