Anyone who know what is the code that i have to write in to get the answer ?????

with the formula : 1+2!/((x-2))-3!/((x-3))+4!/((x-4)) - ... n!/((x-n))
``````#include<iostream>

using namespace std;

int fact (int no)
{
int total;
for (int i=0; i<=no; i++)
{
if(i ==0)
total = 1;
else
total = total * i;
}

}

int main()
{
double x,n;
double result;

cout<<"Please key i x value : ";
cin>>x;
cout<<"\n";
cout<<"Please key in n value : ";
cin>>n;

if (x<n)
{
cout<<"Wrong input, x valur must be greater than n !!\n"<<endl;
}else{
cout<<"Result = 1 + ";
for (int loop = 2; loop< x+1; loop++)
{
if (loop%2 ==0)
{
cout<<loop<<"!/"
<<x-loop;
if (loop == x)
{
cout<<" = ";
}else{
cout<<" - ";
}

}else{
cout<<loop<<"!/"
<<x-loop;
if (loop == x)
cout<<" = ";
else
cout<<" + ";
}
}
}

system("pause");
return 0;
}
``````

As a matter of fact, I do. However, it would not be very helpful for me to simply give the code to you; learning is the goal here, after all.

I will say that you are overthinking things a bit (been there, done that, bought the tee shirt). Rather than having a conditional in which only a single character is changed between the two branches (neither of which gives you what you want), you might consider following the Sigma notation form of the series, and use a sign variable:

``````#include <cmath>      // add this at the top
// ...
int sign = pow(-1, i);
``````

if you use this, and multiply the value (which will be either 1 or -1, depending on the whether the exponent is odd or even) by the result of the factorial over the difference, you can then simply accumulate the result.

``````result += sign * (factorial / denominator);
``````

Another hint is to print the sign along with the value that it refers to, rather than the value before it as you have now.