## C++ program to find the sum of first n natural numbers :

In this *C++ program*, we will learn how to find the sum of first *n* natural numbers. The program will ask the user to enter the value of *n* and it will find the result. We will learn *three* different ways to solve this problem.
This is a beginner level *C++ program* and you will learn how to take user input, how to use loops and how to do mathematical calculations with this program.

### Method 1: By using a for loop:

We can use one *for loop* to iterate from *1* to *n* and find out the sum of all the numbers. Below is the complete program:

```
#include <iostream>
using namespace std;
int main() {
int n;
int sum = 0;
cout << "Enter the value of n :"<<endl;
cin >> n;
for(int i = 1; i<= n; i++){
sum += i;
}
cout << "Total sum :"<<sum<<endl;
return 0;
}
```

*n*is the variable we are storing the user provided value.*sum*variable is used to store the sum. Its initial value is*0*.- The
*for loop*runs from*i = 1*to*i = n*. - Inside the loop, we are adding the current value of
*i*to*sum*. i.e. we at the end of the loop,*sum*holds the sum of*1*to*n*. - Finally, we are printing the total sum i.e.
*sum*

### Sample Output:

```
Enter the value of n :
10
Total sum :55
```

### Method 2: By using a while loop:

Similar to a *for loop*, we can also use one *while loop* to find out the sum. *while* loop checks one condition and runs the loop continuously till the *condition* is true. If we write the above program using *while loop*, it looks as like below:

```
#include <iostream>
using namespace std;
int main() {
int n;
int sum = 0;
int i = 1;
cout << "Enter the value of n :"<<endl;
cin >> n;
while(i <= n){
sum += i;
i++;
}
cout << "Total sum :"<<sum<<endl;
return 0;
}
```

Here, we are defining one variable *i* with value *1*. The *while loop* runs and increment the value of *sum* on each iteration. It prints similar output as the above example.

### Method 3: By using mathematical formula:

The sum of *n* natural numbers is *n*(n + 1)/2*. Instead of using a *loop*, we can directly use this formula. This will be much faster because we donâ€™t have to use the loop. For example, if we have to find the sum of first *1000* numbers, the loop will have to run *1000* times to find out the sum. But if we use the *mathematical formula*, it will find it immediately.

```
#include <iostream>
using namespace std;
int main() {
int n;
int sum;
cout << "Enter the value of n :"<<endl;
cin >> n;
sum = n*(n + 1)/2;
cout << "Total sum :"<<sum<<endl;
return 0;
}
```

It will give similar outputs as the above programs.