Simplification: 1 / (2!) + 2 / (3!) + 3 / (4!) (n-1)/(n!)

What code do you want, C or other languages, I only write the core code to you, you write a main function: 1, recursion: int Cheng (int n) {return n = = 1? 1: n * Cheng (n-1);} This you only need to write a main function, pass the factorial of whose to calculate

It is proved that the order multiplication and division of 2n is n times of 2, and the factorial of n is 1.3.5.7. (2n-1)

A:
(2n)!
=1*2*3*...*(2n-1)*2n
=1*3*5*...*(2n-1)*2*4*6*...*2n
=1*3*5*...*(2n-1)*2^n*(1*2*3*..*n)
=1*3*5*...*(2n-1)*2^n*n!
So (2n)! / (2 ^ n * n!)
=1*3*5*...*(2n-1)*2^n*n!/(2^n*n!)
=1*3*5*...*(2n-1)

Prove: 70! 61! Congruence to module 71

70! - 61! = 70 * 69 * 68 * 67 * 66 * 65 * 64 * 63 * 62 * 61! - 61! = (70 * 69 * 68 * 67 * 66 * 65 * 64 * 63 * 62-1) * 61! Because 70 * 69 * 68 * 67 * 66 * 65 * 64 * 63 * 62 mod 71 = (71-1) (71-2) (71-3) (71-4) (71-5) (71-6) (71-7) (71-8) (71-9) mod 71 = - 9! Mod 71 = - 7! (8 * 9)

C language. Find 1! + 2! + 3! + +Ask to write the function of n factorial

#Include < stdlib. H / x0d ᦇ include < iostream / x0d ᦇ include < iomanip / x0dusing namespace STD; \ x0dconst int n = 1000; \ x0dint compute (unsigned int * s, int n) / / S is used to store the calculation result once, n is the multiplier of this calculation, and the function

Write a program, input positive integer n, count its factorial n! (n! = n × (n-1) ×. × 3 × 2 × 1)

Use until type:
Input "positive integer n = n
I=1
DO
n=n×(n-i)
i=i+1
UNTIL LOOP i=n
Print "factorial =" n "
END

Java number n factorial, factorial, the formula is n! = n * (n-1) (n-2) 2 1. Find the factorial of the factorial of number 6 Java programming

import java.math.BigInteger ;

public class Test {

    public static void main(String[] args) {
        int n = 6;
         System.out.println (jiecheng(n));
    }

    public static BigInteger jiecheng (int i) {
         BigInteger result = new BigInteger("1");
        for (int j = 1; j <= i; j ++) {
               BigInteger num = new BigInteger( String.valueOf (j));
               result = result.multiply (num);
        }
        return result;
    }

Write a program, input positive integer n, calculate its factorial n! (n! = n × (n-1) × ×3×2×1). Don't int fun (int n){ int t=1; Do{ t=t*n; n--; } while(n>=1); return t; }This kind of do not understand, the first chapter algorithm is preliminary. It seems to be to use what VB language

Input "please enter a positive integer"; n
I=n
T=1
DO
t=t*n
i=i-1
LOOPUNTIL i=0
PRINT t
END

C language to find the factorial of N program code

#include
int main()
{
int n,t,i;
t=1;
scanf("%d",&n);
for(i=0;i

C language program: summation of factorials of N, n = 1 to 10

Calculate the sum of factorials from 1 to 10, that is, 1! + 2! + 3! + +Sum of 9! + 10
#include
int main(void)
{
\x09int sum,i,term;
\x09sum=0;
\x09term=1;
\x09for(i=1;i

VB to find the factorial n of n

Private Sub Command1_ Click()
Dim i, j, k
For i = 1 To 10
k = 1
For j = 1 To i
k = k * j
Next
Print i & "!=" & k
Next
End Sub

Simplification: 1 / (2!) + 2 / (3!) + 3 / (4!) (n-1)/(n!)