HMMT 十一月 2009 · 冲刺赛 · 第 21 题

HMMT November 2009 — Guts Round — Problem 21

[ 11 ] Let f ( x ) = x + 2 x + 1. Let g ( x ) = f ( f ( · · · f ( x ))), where there are 2009 f s in the expression for g ( x ). Then g ( x ) can be written as 2009 2009 2 2 − 1 g ( x ) = x + a 2009 x + · · · + a x + a , 1 0 2 − 1 2009 where the a are constants. Compute a . i 2 − 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . nd 2 ANNUAL HARVARD-MIT NOVEMBER TOURNAMENT, 7 NOVEMBER 2009 — GUTS ROUND

解析

[ 11 ] Let f ( x ) = x + 2 x + 1. Let g ( x ) = f ( f ( · · · f ( x ))), where there are 2009 f s in the expression for g ( x ). Then g ( x ) can be written as 2009 2009 2 2 − 1 g ( x ) = x + a 2009 x + · · · + a x + a , 1 0 2 − 1 2009 where the a are constants. Compute a . i 2 − 1 2009 2 n n − 1 n n − 1 2 2 n 2 n − 1 Answer: 2 f ( x ) = ( x +1) , so f ( x + cx + . . . ) = ( x + cx + . . . +1) = x +2 cx + . . . . Applying the preceding formula repeatedly shows us that the coefficient of the term of second highest 2009 degree in the polynomial doubles each time, so after 2009 applications of f it is 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . nd 2 ANNUAL HARVARD-MIT NOVEMBER TOURNAMENT, 7 NOVEMBER 2009 — GUTS ROUND