HMMT 二月 2002 · 团队赛 · 第 5 题

HMMT February 2002 — Team Round — Problem 5

[20] Prove for integers n that ⌊ ⌋ ⌊ ⌋ ⌊ ⌋ 2 n n + 1 n = . 2 2 4 In problems 6–7 you may use without proof the known summations L L ∑ ∑ 3 2 2 n = n ( n + 1) / 2 and n = n ( n + 1) / 4 for positive integers L . n =1 n =1 ∑ L

解析

[20] Prove for integers n that ⌊ ⌋ ⌊ ⌋ ⌊ ⌋ 2 n n + 1 n = . 2 2 4 Solution. Suppose n = 2 m is even; then b n/ 2 c = b m c = m and b ( n + 1) / 2 c = b m + 1 / 2 c = m , ⌊ ⌋ ⌊ ⌋ 2 2 2 whose product is m = m = (2 m ) / 4 . Otherwise n = 2 m + 1 is odd, so that b n/ 2 c = 2 b m + 1 / 2 c = m and b ( n + 1) / 2 c = b m + 1 c = m + 1, whose product is m + m . On the other side, we find that ⌊ ⌋ ⌊ ⌋ ⌊ ⌋ 2 2 n 4 m + 4 m + 1 1 2 2 = = m + m + = m + m, 4 4 4 as desired. In problems 6–7 you may use without proof the known summations L L ∑ ∑ 3 2 2 n = n ( n + 1) / 2 and n = n ( n + 1) / 4 for positive integers L . n =1 n =1 ∑ L