设f(x)=x+1/x-1,记[x]为不超过x的最大整数,那么[f(2)]+[f(3)]+···+[f(2005)]的值是

问题描述:

设f(x)=x+1/x-1,记[x]为不超过x的最大整数,那么[f(2)]+[f(3)]+···+[f(2005)]的值是

当x = 2时候[f(2)] = 3x = 3 时[f(3)] = 2x = 4时[f(4)] = 1当x>=5时[f(x)] = 1所以原式 = 3 +2+ 2002 * 1 = 2007 补充因为如果要满足 (x+1)/(x-1) >=2那么需要 x + 1 >= 2x - 2可以得到x=4以后 [f(x)] = 1...