Problem:
Let P0(x)=x3+313x2−77x−8. For integers n≥1, define Pn(x)=Pn−1(x−n). What is the coefficient of x in P20(x)?
Solution:
For positive integers n, we have
Pn(x)=Pn−1(x−n)=Pn−2(x−n−(n−1))=Pn−3(x−n−(n−1)−(n−2))⋮=P0(x−n−(n−1)−⋯−2−1)
from which
Pn(x)=P0(x−21n(n+1))
Hence
P20(x)=P0(x−2120⋅21)=P0(x−210)=(x−210)3+313(x−210)2−77(x−210)−8.
The coefficient of x in this polynomial is
3(210)2−313⋅2⋅210−77=210(630−626)−77=763
The problems and solutions on this page are the property of the MAA's American Mathematics Competitions