HMMT 二月 2014 · 团队赛 · 第 4 题

HMMT February 2014 — Team Round — Problem 4

[ 20 ] Compute ⌫ 100 100 X 2 . 50 k 2 + 2 k =0 (Here, if x is a real number, then b x c denotes the largest integer less than or equal to x .)

解析

[ 20 ] Compute ⌊ ⌋ 100 100 ∑ 2 . 50 k 2 + 2 k =0 (Here, if x is a real number, then ⌊ x ⌋ denotes the largest integer less than or equal to x .) 100 2 49 Answer: 101 · 2 − 50 Let a = . Notice that, for k = 0 , 1 , . . . , 49, k 50 k 2 +2 100 100 100 50+ k 2 2 2 2 50 a + a = + = + = 2 k 100 − k 50 k 50 100 − k 50 k k 50 2 + 2 2 + 2 2 + 2 2 + 2 50 It is clear that for k = 0 , 1 , . . . , 49, a , a ∈ / Z , so ⌊ a ⌋ + ⌊ a ⌋ = 2 − 1 (since the sum of floors k 100 − k k 100 − k is an integer less than a + a but greater than a − 1 + a − 1). Thus, k 100 − k k 100 − k 100 ∑ 50 49 49 ⌊ a ⌋ = 50 · (2 − 1) + 2 = 101 · 2 − 50 k k =0