HMMT 二月 2006 · CALC 赛 · 第 4 题
HMMT February 2006 — CALC Round — Problem 4
题目详情
- Compute . k ! k =1 ∫ 1 dx
解析
- Compute . k ! k =1 Answer: 15 e Solution: Define, for non-negative integers n , ∞ n ∑ k S := , n k ! k =0 1 0 where 0 = 1 when it occurs. Then S = e , and, for n ≥ 1, 0 ( ) ∞ ∞ ∞ ∞ n − 1 n n n n − 1 ∑ ∑ ∑ ∑ ∑ k k ( k + 1) ( k + 1) n − 1 S = = = = = S , n i k ! k ! ( k + 1)! k ! i i =0 k =0 k =1 k =0 k =0 so we can compute inductively that S = e , S = 2 e , S = 5 e , and S = 15 e . 1 2 3 4