C++ Program to Find Sum of N Natural Numbers

To find the sum of n natural numbers in C++, use the formula sum = n * (n + 1) / 2 or a loop to add numbers from 1 to n. For example: int sum = n * (n + 1) / 2; calculates the sum directly.

Examples

Input5

OutputSum of first 5 natural numbers is 15

Input10

OutputSum of first 10 natural numbers is 55

Input1

OutputSum of first 1 natural number is 1

How to Think About It

To find the sum of the first n natural numbers, think of adding all numbers starting from 1 up to n. You can do this by adding each number one by one using a loop or use the simple math formula n * (n + 1) / 2 which gives the total sum directly without looping.

Algorithm

1

Get input value n from the user

2

Calculate sum using the formula sum = n * (n + 1) / 2

3

Print the sum

Code

cpp

#include <iostream>
using namespace std;

int main() {
    int n;
    cout << "Enter a positive integer: ";
    cin >> n;
    int sum = n * (n + 1) / 2;
    cout << "Sum of first " << n << " natural numbers is " << sum << endl;
    return 0;
}

Output

Enter a positive integer: 5
Sum of first 5 natural numbers is 15

Dry Run

Let's trace input 5 through the code

1

Input

User enters n = 5

2

Calculate sum

sum = 5 * (5 + 1) / 2 = 5 * 6 / 2 = 30 / 2 = 15

3

Output

Print 'Sum of first 5 natural numbers is 15'

n	sum
5	15

Why This Works

Step 1: Input value

We get the number n from the user which tells how many natural numbers to add.

Step 2: Use formula

The formula n * (n + 1) / 2 quickly calculates the sum without looping.

Step 3: Display result

We print the sum so the user sees the total of the first n natural numbers.

Alternative Approaches

Using a loop

cpp

#include <iostream>
using namespace std;

int main() {
    int n, sum = 0;
    cout << "Enter a positive integer: ";
    cin >> n;
    for (int i = 1; i <= n; ++i) {
        sum += i;
    }
    cout << "Sum of first " << n << " natural numbers is " << sum << endl;
    return 0;
}

This method uses a loop to add each number one by one, which is easy to understand but slower for very large n.

Using recursion

cpp

#include <iostream>
using namespace std;

int sumNatural(int n) {
    if (n == 1) return 1;
    return n + sumNatural(n - 1);
}

int main() {
    int n;
    cout << "Enter a positive integer: ";
    cin >> n;
    cout << "Sum of first " << n << " natural numbers is " << sumNatural(n) << endl;
    return 0;
}

This method uses recursion to add numbers but can cause stack overflow for very large n.

Complexity: O(1) time, O(1) space

Time Complexity

Using the formula calculates the sum in constant time because it performs only a few arithmetic operations.

Space Complexity

The program uses a fixed amount of memory regardless of input size, so space complexity is constant.

Which Approach is Fastest?

The formula method is fastest and most efficient compared to looping or recursion which take O(n) time.

Approach	Time	Space	Best For
Formula	O(1)	O(1)	Fastest calculation for any n
Loop	O(n)	O(1)	Simple logic, easy to understand
Recursion	O(n)	O(n)	Demonstrates recursion but less efficient