HMMT 十一月 2009 · 冲刺赛 · 第 21 题
HMMT November 2009 — Guts Round — Problem 21
题目详情
- [ 11 ] Let f ( x ) = x + 2 x + 1. Let g ( x ) = f ( f ( · · · f ( x ))), where there are 2009 f s in the expression forg ( 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
2 ANNUAL HARVARD-MIT NOVEMBER TOURNAMENT, 7 NOVEMBER 2009 — GUTS ROUNDnd
英文原题
[ 11 ] Let f ( x ) = x 2 + 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
g ( x ) = x 2 2009
- a 2 2009 − 1 x 2 2009 − 1 + · · · + a 1 x + a 0 ,
where the a i are constants. Compute a 2 2009 − 1 .
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2 nd ANNUAL HARVARD-MIT NOVEMBER TOURNAMENT, 7 NOVEMBER 2009 — GUTS ROUND
解析
英文解析
- [ 11 ] Let f ( x ) = x + 2 x + 1. Let g ( x ) = f ( f ( · · · f ( x ))), where there are 2009 f s in the expression forg ( 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 .
2 ANNUAL HARVARD-MIT NOVEMBER TOURNAMENT, 7 NOVEMBER 2009 — GUTS ROUNDnd