Difference between revisions of "1993 AIME Problems/Problem 5"
m |
(solution) |
||
| Line 1: | Line 1: | ||
== Problem == | == Problem == | ||
| − | Let <math>P_0(x) = x^3 + 313x^2 - 77x - 8\,</math>. For | + | Let <math>P_0(x) = x^3 + 313x^2 - 77x - 8\,</math>. For [[integer]]s <math>n \ge 1\,</math>, define <math>P_n(x) = P_{n - 1}(x - n)\,</math>. What is the [[coefficient]] of <math>x\,</math> in <math>P_{20}(x)\,</math>? |
== Solution == | == Solution == | ||
| − | {{ | + | Notice that <math>\displaystyle P_{20}(x) = P_{19}(x - 20) = P_{18}((x - 20) - 19)</math><math>\displaystyle = P_{17}(((x - 20) - 19) - 18) \ldots</math><math>\displaystyle = P_0(x - (20 + 19 + 18 + \ldots + 2 + 1))</math>. Using the formula for the sum of the first <math>n</math> numbers, <math>1 + 2 \ldots + 20 = \frac{20(20+1)}{2} = 210</math>. Thus, <math>\displaystyle P_{20}(x) = P_0(x - 210)</math>. |
| + | |||
| + | Substitute <math>\displaystyle x - 210</math> into the equation, so we get <math>\displaystyle (x - 210)^3 + 313(x - 210)^2 - 77(x - 210) - 8</math>. The cubic will have a term of <math>{3\choose1}210^2x = 630 \cdot 210x</math>. The square will have a term of <math>-313 \cdot {2\choose1}210x = -626 \cdot 210x</math>. The linear part will have a term of <math>\displaystyle -77x</math>. Adding up the coefficients, we get <math>630 \cdot 210 - 626 \cdot 210 - 77 = 763</math>. | ||
== See also == | == See also == | ||
{{AIME box|year=1993|num-b=4|num-a=6}} | {{AIME box|year=1993|num-b=4|num-a=6}} | ||
| + | |||
| + | [[Category:Intermediate Algebra Problems]] | ||
Revision as of 19:02, 27 March 2007
Problem
Let 
. For integers 
, define 
. What is the coefficient of 
in 
?
Solution
Notice that 
. Using the formula for the sum of the first 
numbers, 
. Thus, 
.
Substitute 
into the equation, so we get 
. The cubic will have a term of 
. The square will have a term of 
. The linear part will have a term of 
. Adding up the coefficients, we get 
.
See also
| 1993 AIME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 4 |
Followed by Problem 6 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
