**Problem statement: **Given a number ‘N’, find out the sum of the first N natural numbers.

**Examples:**

Example 1:Input:N=5Output:15Explanation:1+2+3+4+5=15Example 2:Input:N=6Output:21Explanation:1+2+3+4+5+6=15

**Solution**

*Disclaimer*: *Don’t jump directly to the solution, try it out yourself first.*

**Solution1: Using Loop**

**Intuition: **We can simply add numbers one by one from 1 to N.

For eg. if N = 5, we can add 1+2+3+4+5=15.

We can use a for loop or while loop to achieve the goal.

**Approach: **

- Take a variable sum and initialize it as 0.
- Take a for loop and run from 1 to N.
- Save the result in sum.

**Code:**

## C++ Code

```
#include<bits/stdc++.h>
using namespace std;
void solve(int n) {
int sum = 0;
for (int i = 1; i <= n; i++) {
sum += i;
}
cout<<"The sum of the first "<<n<<" numbers is: "<<sum<<endl;
}
int main() {
solve(5);
solve(6);
}
```

**Output:**

The sum of the first 5 numbers is: 15

The sum of the first 6 numbers is: 21

**Time Complexity:** O(N)

**Space Complexity:** O(1)

## Java Code

```
import java.util.*;
public class tuf {
public static void main(String[] args) {
solve(5);
solve(6);
}
public static void solve(int n) {
int sum = 0;
for (int i = 1; i <= n; i++) {
sum += i;
}
System.out.println("The sum of the first " + n + " numbers is: " + sum);
}
}
```

**Output:**

The sum of the first 5 numbers is: 15

The sum of the first 6 numbers is: 21

**Time Complexity:** O(N)

**Space Complexity:** O(1)

**Solution 2: Using the formula**

**Intuition: **We can use the formula for the sum of N numbers, i.e** N(N+1)/2.**

For eg: N=5

5(5+1)/2 = 5(6)/2 = 15.

**Approach: **

- Take a variable sum.
- Initialize it with
**N(N+1)/2,**where N is a given number.

**Code:**

## C++ Code

```
#include<bits/stdc++.h>
using namespace std;
void solve(int N) {
int sum = N * (N + 1) / 2;
cout<<"The sum of the first "<<N<<" numbers is: "<<sum<<endl;
}
int main() {
solve(5);
solve(6);
}
```

**Output:**

The sum of the first 5 numbers is: 15

The sum of the first 6 numbers is: 21

**Time Complexity:** O(1)

**Space Complexity:** O(1)

## Java Code

```
import java.util.*;
public class tuf {
public static void main(String[] args) {
solve(5);
solve(6);
}
public static void solve(int N) {
int sum = N * (N + 1) / 2;
System.out.println("The sum of the first " + N + " numbers is: " + sum);
}
}
```

**Output:**

The sum of the first 5 numbers is: 15

The sum of the first 6 numbers is: 21

**Time Complexity:** O(1)

**Space Complexity:** O(1)

Special thanks toplease check out this articlePrashant Sahufor contributing to this article on takeUforward. If you also wish to share your knowledge with the takeUforward fam,