HMMT 二月 2007 · TEAM1 赛 · 第 14 题

HMMT February 2007 — TEAM1 Round — Problem 14

[ 40 ] Find an explicit, closed form formula for ( ) n n k ∑ k · ( − 1) · k . n + k + 1 k =1 2

解析

[ 40 ] Find an explicit, closed form formula for ( ) n n k ∑ k · ( − 1) · k . n + k + 1 k =1 n !( n + 1 )! − 1 Answer: or − or obvious equivalent . 2n + 1 ( 2n + 1 )! ( ) n Solution. Consider the interpolation of the polynomial P ( x ) = x · n ! at x = 0 , 1 , . . . , n . We obtain the identity n ∑ ∏ x − j P ( x ) = x · n ! = k · n ! k − j k =0 j 6 = k n ∑ x ( x − 1) · · · ( x − k + 1)( x − k − 1) · · · ( x − n ) = k · n ! · n − k k !( n − k )!( − 1) k =0 ( ) n ∑ n n − k = k · ( − 1) · · x ( x − 1) · · · ( x − k + 1)( x − k − 1) · · · ( x − n ) . k k =1 1 This identity is valid for all complex numbers x , but, to extract a factor from the valid product n + k +1 of each summand, we set x = − n − 1, so that ( ) ( ) n n n k ∑ ∑ k ( − 1) (2 n + 1)! n n − k k − ( n +1)! = k ( − 1) ( − n − 1) · · · ( − n − k )( − n − k − 2) · · · ( − 2 n − 1) = . k n !( n + k + 1) k =1 k =1 Finally, ( ) n n k ∑ k · ( − 1) · − n !( n + 1)! − 1 k = = ( ) . 2 n +1 n + k + 1 (2 n + 1)! n k =1 6