HMMT 二月 2019 · ALGNT 赛 · 第 7 题
HMMT February 2019 — ALGNT Round — Problem 7
题目详情
- Find the value of ∞ ∞ ∞ ∑ ∑ ∑ ab (3 a + c ) . a + b + c 4 ( a + b )( b + c )( c + a ) a =1 b =1 c =1
解析
- Find the value of ∞ ∞ ∞ ∑ ∑ ∑ ab (3 a + c ) . a + b + c 4 ( a + b )( b + c )( c + a ) a =1 c =1 b =1 Proposed by: Andrew Gu 1 Answer: 54 Let S denote the given sum. By summing over all six permutations of the variables a, b, c we obtain ∞ ∞ ∞ 2 2 2 2 2 2 ∑ ∑ ∑ 3( a b + a c + b a + b c + c a + c b ) + 6 abc 6 S = a + b + c 4 ( a + b )( b + c )( c + a ) a =1 b =1 c =1 ∞ ∞ ∞ ∑ ∑ ∑ 3 = a + b + c 4 a =1 c =1 b =1 ( ) ( ) ( ) ∞ ∞ ∞ ∑ ∑ ∑ 1 1 1 = 3 a b c 4 4 4 a =1 c =1 b =1 ( ) 3 1 = 3 3 1 = . 9 1 Hence S = . 54