算法-计算n的阶乘以及n的累加

编写算法，计算下面表达式的值：
1 + （1+2 + 1 * 2）+ (1+2+3 + 1 * 2 * 3) +…+ (1+2+…+n + 1 * 2 *…*n)；

public class AutoTest {
    public static Integer sum;
    /**
     * 编写算法，计算下面表达式的值
     * 1 + （1+2 + 1*2）+ (1+2+3 + 1*2*3) +...+ (1+2+...+n + 1*2*...*n)
     * 1
     * 1 + 5 = 6
     * 1 + 5 + 12 = 18
     * 1 + 5 + 12 + 34 = 52
     * 1 + 5 + 12 +34 + 135 = 187
     */
    public static Integer countSum(int n){
        sum = 0;
        for (int i = 1; i <= n; i++) {
            sum += i * (n - i + 1) + (i == 1 ? 0:recursion(i));
        }
        return sum;
    }
    public static Integer recursion(int n){
        if(n == 1){
            return 1;
        }
        return n*recursion(n-1);
    }

     public static void main(String[] args) {
        System.out.println("3     :" + countSum(3));
        System.out.println("4     :" + countSum(4));
        System.out.println("5     :" + countSum(5));
        System.out.println("6     :" + countSum(6));
        System.out.println("7     :" + countSum(7));
    }
}

验证结果：

3     :18
4     :52
5     :187
6     :928
7     :5996