已知数列{an}的通项an=f(n)(n∈N+),求an的最大值或最小值.对于这个问题,目前普遍的解法是通过如下不等式组来确定取最大值或最小值时n的值即:要使an最大,则应满足an≥an-1,an≥an+1,其中n≥2.要使an最小,则应满足an≤an-1,an≤an+1,其中n≥2.通过以上的不等式组求解出n的值,进 Knowing the general term an = f(n) (n∈N+) of the series {an}, finds the maximum or minimum value of an. For this problem, the current general solution is to determine the maximum or minimum by the following set of inequalities. When the value is n, the value is: To make an maximum, we should satisfy an≥an-1,an≥an+1, where n≥2. To make an minimum, we should satisfy an≤an-1,an≤an+ 1, where n ≥ 2. Solve the value of n through the above inequality group, enter