因为{a n }是等差数列,若n+1-(m-1)=n-m+2为偶数,根据等差中项的概念, 则由a m +a m+1 +…+a n+1 =0,得: n-m+2 2 ( a m + a n+1 )=0 ,因为 n-m+2 2 ≠0 ,所以a m +a n+1 =0. 若n+1-(m-1)=n-m+2为奇数, 则由a m +a m+1 +…+a n+1 =0,得: n-m+1 2 ( a m + a n+1 )+ 1 2 ( a m + a n+1 ) = n-m+2 2 ( a m + a n+1 )=0 , 因为 n-m+2 2 ≠0 ,所以a m +a n+1 =0. 又a 1 +a m+n =a m +a n+1 , 则 S m+n = ( a 1 + a m+n )n 2 = ( a m + a n+1 )n 2 =0 . 故选C.