已知阶乘结果求阶数 求7的阶乘

求7的阶乘  

求7的阶乘

C#控制台：
using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Threading.Tasks;
namespace ConApp
{
class Program
{
static void Main(string[] args)
{
Console.WriteLine("请输入一个数字，并计算改数字的阶乘：");
int n = Convert.ToInt32(Console.ReadLine());
for (int i = n-1; i > 0; i--)
{
n = n * i;
}
Console.WriteLine(n);
Console.ReadKey();
}
}
}
7阶乘等于5040

C语言 5×5阶乘+6×5阶乘+7×5阶乘+8×5阶乘

#include"stdio.h"
void main()
{
int i,j,t,sum;
t=1;sum=0;
for (i=2;i<=5;i++)
t*=i;
for (i=5;i<=8;i++)
sum+=i*t;
printf("%dn",sum);
}

1到7阶乘阶乘和为

public class a {
public static void main(String[] args) {
long a=1;long sum=1;long allsum=0;

while(a<2){
allsum+=sum;
long b=a;
a=++a;
sum=a*b;
}
System.out.println(sum);
}
}

用java语言编写1的阶乘减2的阶乘加3的阶乘减4的阶乘加5的阶乘减6的阶乘

sum = 0;
addOrMinus=1;
for(int i=1;i<=6;i++){
sum += i!*addOrMinus;
addOrMinus *= -1;
}
System.out.println(sum);

用pascal语言求出 1的阶乘+2的阶乘+···+n的阶乘的值

function p(k:longint):qword;
var i:longint;
begin
p:=1;
for i:=2 to k do {本来是for i:=1 to k do的，为了少n次回圈，改为for i:=2 to k do}
p:=p*i;
end;
var m,n:longint;
s:qword;
begin
read(n);
s:=0;
for m:=1 to n do
s:=s+p(m);
writeln(s);
end.

求7 8 9 10的阶乘，vb语言

Private Sub Combo1_Click()
 Dim s As String
    Dim n As Integer
    n = Combo1.Text
    s = ""
    s = "计算1!+2!+3!+...+" & n & "!,其结果为" & jiecheng(n)
    Text1.Text = s
End Sub
Private Sub Form_Load()
Combo1.Clear
Combo1.AddItem ("7")
Combo1.AddItem ("8")
Combo1.AddItem ("9")
Combo1.AddItem ("10")
End Sub
Public Function jiecheng(ByVal n As Integer)
        Dim m As Double
    m = 1
        For i = 1 To n
            m = m * i
        Next
jiecheng = m
    End Function

1的阶乘+2倍2的阶乘+.......+n倍n的阶乘=?

n*n!=[(n+1)-1]*n!=(n+1)!-n!
1*1!=2!-1!
2*2!=3!-2!
……
1的阶乘+2倍2的阶乘+.......+n倍n的阶乘=(2!-1!) +(3!-2!) +(4!-3!) +……+[(n+1)!-n!]=(n+1)!-1!=(n+1)!-1

求（1/2的阶乘+2/3的阶乘+。。。+n/(n+1)的阶乘）的极限

n/(n+1)!=1/n!-1/(n+1)!,
（1/2的阶乘+2/3的阶乘+。。。+n/(n+1)的阶乘）=1/n!-1/(n+1)!+1/(n-1)!-1/n!+...
+1/2!-1/3!+1/1!-1/2!=1-1/(n+1)!
故（1/2的阶乘+2/3的阶乘+。。。+n/(n+1)的阶乘）的极限为1.

1的阶乘+2的阶乘+3的阶乘+``+100的阶乘的个位数字是( )

5和5以上的数的阶乘中有5*2,所以个位一定为0.只要算1!+2!+3!+4!=1+2+2*3+2*3*4=1+2+6+24=33 因此个位为3