HMMT 二月 2013 · 代数 · 第 7 题

HMMT February 2013 — Algebra — Problem 7

Compute ∞ ∞ ∞ ∑ ∑ ∑ a + a + · · · + a 1 2 7 · · · . a + a + ··· + a 1 2 7 3 a =0 a =0 a =0 1 2 7 2 2 4 4 6 6 x + y x + y x + y

解析

Compute ∞ ∞ ∞ ∑ ∑ ∑ a + a + · · · + a 1 2 7 · · · . a + a + ··· + a 1 2 7 3 a =0 a =0 a =0 1 2 7 Answer: 15309/256 Note that, since this is symmetric in a through a , 1 7 ∞ ∞ ∞ ∞ ∞ ∞ ∑ ∑ ∑ ∑ ∑ ∑ a + a + · · · + a a 1 2 7 1 · · · = 7 · · · a + a + ··· + a a + a + ··· + a 1 2 7 1 2 7 3 3 a =0 a =0 a =0 a =0 a =0 a =0 1 2 7 1 2 7 ( ) ( ) 6 ∞ ∞ ∑ ∑ a 1 1 = 7 . a a 1 3 3 a =0 a =0 1 ( ) ∑ ∑ 6 a 1 3 3 15309 If S = , then 3 S − S = = 3 / 2, so S = 3 / 4. It follows that the answer equals 7 · · = . a a 3 3 4 2 256 ∑ ∑ ∑ ∞ ∞ ∞ a + a + ··· + a 1 2 7 Alternatively, let f ( z ) = · · · z . Note that we can rewrite f ( z ) = a =0 a =0 a =0 1 2 7 ∑ ∑ ∑ ∑ ∞ ∞ ∞ ∞ a 7 1 ′ ( z ) = . Furthermore, note that zf ( z ) = · · · ( a + a + · · · + 7 1 2 a =0 a =0 a =0 a =0 (1 − z ) 1 2 7 ′ f (1 / 3) 7 a + a + ··· + a ′ 1 2 7 a ) z , so the sum in question is simply . Since f ( x ) = , it follows that the sum 7 8 3 (1 − z ) 7 7 · 3 15309 is equal to = . 8 2 256 2 2 4 4 6 6 x + y x + y x + y