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−21​n(n+1))
Hence
P20​(x)=P0​(x−21​20⋅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 on this page are the property of the MAA's American Mathematics Competitions