an = 2 - 1 / a(n-1) ,形如这样的数列可以如下计算：因为 an + k = k + 2 - 1 / a(n-1) = [(k + 2) a(n-1) - 1] / a(n-1) ,所以 1 / (an + k) = a(n-1) / [(k + 2) a(n-1) - 1] 令 k (k + 2) = - 1 ,得 k^2 + 2 k + ...