HMMT 二月 2002 · 代数 · 第 10 题

HMMT February 2002 — Algebra — Problem 10

Determine the value of 1 1 1 1 2002 + (2001 + (2000 + · · · + (3 + · 2)) · · · ) . 2 2 2 2 1

解析

Determine the value of 1 1 1 1 2002 + (2001 + (2000 + · · · + (3 + · 2)) · · · ) . 2 2 2 2 1 1 1 Solution: 4002 . We can show by induction that n + ([ n − 1]+ ( · · · + · 2) · · · ) = 2( n − 1). 2 2 2 1 For n = 3 we have 3 + · 2 = 4, giving the base case, and if the result holds for n , then 2 1 ( n + 1) + 2( n − 1) = 2 n = 2( n + 1) − 2. Thus the claim holds, and now plug in n = 2002. 2 2 2000 Alternate Solution: Expand the given expression as 2002+2001 / 2+2000 / 2 + · · · +2 / 2 . 2 2001 Letting S denote this sum, we have S/ 2 = 2002 / 2 + 2001 / 2 + · · · + 2 / 2 , so S − S/ 2 = 2000 2001 2000 2000 2002 − (1 / 2 + 1 / 4 + · · · + 1 / 2 ) − 2 / 2 = 2002 − (1 − 1 / 2 ) − 1 / 2 = 2001, so S = 4002. 3