Need to calculate series equation converging to π - c++

Question

First time programming and I’ve got to write a program that calculates the series equation
(n!)^2*2^n+1/(2n+1)!, n being the amount of terms the user inputs.
I’ve got it to where the user inputs n and I get the answer for that number only.
How do I make it so that I get the sum of all answers from 0 to the user input?
#include <iostream>
#include<cmath>
using namespace std;
int main()
{
double i,n,factorial,factorial2,n2,a;
a = 1;
cout<<"Enter # of terms:";
cin>>n;
for (i = 0; i <= n; i++)
if (i == 0)
factorial = 1;
else
factorial = factorial * i;
factorial = pow(factorial,2)*pow(2,n+1);
n2 = 2*n+a;
for (i = 0; i <= n2; i++)
if (i == 0)
factorial2 = 1;
else
factorial2 = factorial2 * i;
factorial = factorial/factorial2;
cout<<factorial<<endl;
system("PAUSE");
}

Answer 1

Here's the code from my comments, all laid out to be readable...
#include <iostream>
#include <cmath>
int factorial(int n)
{
return n <= 1 ? 1 : factorial(n - 1) * n;
}
OR if you find that factorial function confusing, SlySherZ suggested this beginner-friendly alternative (you don't want both implementations).
int factorial(int n)
{
if (n <= 1)
return 1;
return factorial(n - 1) * n;
}
Continuing...
double f(double n)
{
return std::pow(fact(n),2) * std::pow(2, n) + 1 / factorial(2 * n + 1);
}
int main()
{
double n, total = 0;
while (std::cin >> n)
total += f(n);
std::cout << "total " << total << '\n';
}

Answer 2

You have two options:
Use a loop that will go through each n;
Make a recursive function:
// In pseudocode
term (n){
if ( n < 0 ) return 0;
// calculate y for this n;
return y + term( n - 1 );
}
A recursive function (assuming you're familiar with functions) is a function that calls itself to solve a problem. For example, n! = (n - 1)! * n. You can use this simple fact an create a factorial function f ( n ) that returns n * f ( n - 1 ) unless n is 1 (returns 1 in this case).
I'll develop the first option:
// I assumed your code is working and that factorial holds the values you want to sum.
// If that's not the case, I believe you can use it to solve your problem anyway
#include <iostream>
#include<cmath>
using namespace std;
int main() {
double n, total = 0;
cout << "Enter # of terms:";
cin >> n;
while (n >= 0) {
double i, factorial, factorial2, n2, a;
a = 1;
for (i = 0; i <= n; i++) {
if (i == 0)
factorial = 1;
else
factorial = factorial * i;
}
factorial = pow(factorial, 2)*pow(2, n + 1);
n2 = 2 * n + a;
for (i = 0; i <= n2; i++) {
if (i == 0)
factorial2 = 1;
else
factorial2 = factorial2 * i;
factorial = factorial / factorial2;
}
total += factorial;
n--;
}
cout << total << endl;
}
Hope it helps :D