解析:
前n项和为S=

+

+

+……+

=

×(

-

+

-

-

+……+

-

)=

×(

-

)=

,故其极限为

,所以选A。
有以下程序
#include
#include
typedef struct { char name[9]; char sex; float score[2]; } STU;
void f(STU A)
{
STU b={"Zhao", 'm', 85.0, 90.0};
int i;
strcpy(a.name, b.name);
a.sex = b.sex;
for (i=0; i<2; i++)
a.score[i] = b.score[i];
}
main()
{
STU c={"Qian", 'f', 95.0, 92.0};
f(c);
printf("%s,%c,%2.0f,%2.0f\n", c.name, c.sex, c.score[0], c.score[1]);
}
程序的运行结果是
若有以下程序
#include
main()
{ char w[20], a[5][10]={"abcdef", "ghijkl", "mnopq", "rstuv", "wxyz"};
int p[6][2]={{0,1},{1,5},{0,0},{0,2},{1,4},{4,4}}, i;
for (i=0; i<6; i++) w[i]=a[ p[i][0] ][ p[i][1] ];
puts(w);
}
则程序的输出结果是
在窗体上画一个名称为Command1的命令按钮 ,并编写如下程序:Option Base 1Private Sub Command1_Click() Dim a(4, 4) For i=1 To 4 For j=1 To 4 a(i, j)=(i-1)* 3+j Next j Next i For i=3 To 4 For j=3 To 4 Print a(j, i); Next j Print Next iEnd Sub运行程序,单击命令按钮,则输出结果为( )。