有一列数a1,a2,a3,.,an,从第二个数开始,每一个数都等于1于它前面的数的倒数的差,若a1=2,则a2009=

a[n] = 1 - 1/a[n-1]= (a[n-1]-1)/a[n-1]a[n+1] = 1 - 1/a[n]= 1 - a[n-1]/(a[n-1]-1)= - 1/(a[n-1]-1)a[n+2] = 1 - 1/a[n+1]= a[n-1].∴ a[n+3] = a[n]∴ a[2009] = a[2] = 1 - 1/a[1] = 1/2.应该没有错吧``` ....