# Program to find sum of first n natural numbers

Given a number n, find the sum of first natural numbers.

Examples :

```Input : n = 3
Output : 6
Explanation :
Note that 1 + 2 + 3 = 6

Input  : 5
Output : 15
Explanation :
Note that 1 + 2 + 3 + 4 + 5 = 15
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

A simple solution is to do following.

```1) Initialize : sum = 0
2) Run a loop from x = 1 to n and
do following in loop.
sum = sum + x ```
 `// CPP program to find sum of first ` `// n natural numbers. ` `#include ` `using` `namespace` `std; ` ` `  `// Returns sum of first n natural ` `// numbers ` `int` `findSum(``int` `n) ` `{ ` `   ``int` `sum = 0; ` `   ``for` `(``int` `x=1; x<=n; x++)  ` `     ``sum = sum + x; ` `   ``return` `sum; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `  ``int` `n = 5; ` `  ``cout << findSum(n); ` `  ``return` `0; ` `}  `

 `// JAVA program to find sum of first ` `// n natural numbers. ` `import` `java.io.*; ` ` `  `class` `GFG{ ` ` `  `    ``// Returns sum of first n natural ` `    ``// numbers ` `    ``static` `int` `findSum(``int` `n) ` `    ``{ ` `        ``int` `sum = ``0``; ` `        ``for` `(``int` `x = ``1``; x <= n; x++)  ` `            ``sum = sum + x; ` `        ``return` `sum; ` `    ``} ` ` `  `    ``// Driver code ` `    ``public` `static` `void` `main(String args[]) ` `    ``{ ` `        ``int` `n = ``5``; ` `        ``System.out.println(findSum(n)); ` `    ``}  ` `} ` ` `  `// This code is contributed by Nikita Tiwari. `

 `# PYTHON program to find sum of first ` `# n natural numbers. ` ` `  `# Returns sum of first n natural ` `# numbers ` `def` `findSum(n) : ` `    ``sum` `=` `0` `    ``x ``=` `1` `    ``while` `x <``=``n : ` `        ``sum` `=` `sum` `+` `x ` `        ``x ``=` `x ``+` `1` `    ``return` `sum` ` `  ` `  `# Driver code ` ` `  `n ``=` `5` `print` `findSum(n) ` ` `  `# This code is contributed by Nikita Tiwari. `

 `// C# program to find sum of first ` `// n natural numbers. ` `using` `System; ` ` `  `class` `GFG{ ` ` `  `    ``// Returns sum of first n natural ` `    ``// numbers ` `    ``static` `int` `findSum(``int` `n) ` `    ``{ ` `        ``int` `sum = 0; ` `        ``for` `(``int` `x = 1; x <= n; x++)  ` `            ``sum = sum + x; ` `        ``return` `sum; ` `    ``} ` ` `  `    ``// Driver code ` `    ``public` `static` `void` `Main() ` `    ``{ ` `        ``int` `n = 5; ` `        ``Console.Write(findSum(n)); ` `    ``}  ` `} ` ` `  `// This code is contributed by vt_m. `

 ` `

Output :
```15
```

An efficient solution is to use below formula.

How does this work?

```We can prove this formula using induction.

It is true for n = 1 and n = 2
For n = 1, sum = 1 * (1 + 1)/2 = 1
For n = 2, sum = 2 * (2 + 1)/2 = 3

Let it be true for k = n-1.

Sum of k numbers = (k * (k+1))/2
Putting k = n-1, we get
Sum of k numbers = ((n-1) * (n-1+1))/2
= (n - 1) * n / 2

If we add n, we get,
Sum of n numbers = n + (n - 1) * n / 2
= (2n + n2 - n)/2
= n * (n + 1)/2
```
 `// Efficient CPP program to find sum of first ` `// n natural numbers. ` `#include ` `using` `namespace` `std; ` ` `  `// Returns sum of first n natural ` `// numbers ` `int` `findSum(``int` `n) ` `{ ` `   ``return` `n * (n + 1) / 2; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `  ``int` `n = 5; ` `  ``cout << findSum(n); ` `  ``return` `0; ` `}  `

 `// Efficient JAVA program to find sum  ` `// of first n natural numbers. ` `import` `java.io.*; ` ` `  `class` `GFG{ ` `     `  `    ``// Returns sum of first n natural ` `    ``// numbers ` `    ``static` `int` `findSum(``int` `n) ` `    ``{ ` `        ``return` `n * (n + ``1``) / ``2``; ` `    ``} ` ` `  `    ``// Driver code ` `    ``public` `static` `void` `main(String args[]) ` `    ``{ ` `        ``int` `n = ``5``; ` `        ``System.out.println(findSum(n)); ` `    ``} ` `} ` ` `  `// This code is contributed by Nikita Tiwari. `

 `# Efficient CPP program to find sum  ` `# of first n natural numbers. ` ` `  `# Returns sum of first n natural ` `# numbers ` `def` `findSum(n) : ` `    ``return` `n ``*` `(n ``+` `1``) ``/` `2` `     `  `# Driver code ` `n ``=` `5` `print` `findSum(n) ` ` `  `# This code is contributed by Nikita Tiwari. `

 `// Efficient C# program to find sum  ` `// of first n natural numbers. ` `using` `System; ` ` `  `class` `GFG{ ` `     `  `    ``// Returns sum of first n natural ` `    ``// numbers ` `    ``static` `int` `findSum(``int` `n) ` `    ``{ ` `        ``return` `n * (n + 1) / 2; ` `    ``} ` ` `  `    ``// Driver code ` `    ``public` `static` `void` `Main() ` `    ``{ ` `        ``int` `n = 5; ` `        ``Console.Write(findSum(n)); ` `    ``} ` `} ` ` `  `// This code is contributed by vt_m. `

 ` `

Output:
```15
```

The above program causes overflow, even if the result is not beyond integer limit
. We can avoid overflow up to some extent by doing division first.

 `// Efficient CPP program to find sum of first ` `// n natural numbers that avoids overflow if ` `// result is going to be within limits. ` `#include ` `using` `namespace` `std; ` ` `  `// Returns sum of first n natural ` `// numbers ` `int` `findSum(``int` `n) ` `{ ` `   ``if` `(n % 2 == 0) ` `      ``return` `(n/2) * (n+1); ` ` `  `   ``// If n is odd, (n+1) must be even ` `   ``else`  `      ``return`  `((n + 1) / 2) * n; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `  ``int` `n = 5; ` `  ``cout << findSum(n); ` `  ``return` `0; ` `}  `

 `// Efficient JAVA program to find sum of first ` `// n natural numbers that avoids overflow if ` `// result is going to be within limits. ` `import` `java.io.*; ` ` `  `class` `GFG{ ` ` `  `    ``// Returns sum of first n natural ` `    ``// numbers ` `    ``static` `int` `findSum(``int` `n) ` `    ``{ ` `        ``if` `(n % ``2` `== ``0``) ` `            ``return` `(n / ``2``) * (n + ``1``); ` ` `  `        ``// If n is odd, (n+1) must be even ` `        ``else` `            ``return` `((n + ``1``) / ``2``) * n; ` `    ``} ` ` `  `    ``// Driver code ` `    ``public` `static` `void` `main(String args[]) ` `    ``{ ` `        ``int` `n = ``5``; ` `        ``System.out.println(findSum(n)); ` `    ``} ` `} ` ` `  `//This code is contributed by Nikita Tiwari. `

 `# Efficient Python program to find the sum   ` `# of first n natural numbers that avoid  ` `# overflow if the result is going to be  ` `# within limits. ` ` `  `# Returns sum of first n natural ` `# numbers ` `def` `findSum(n) : ` `    ``if` `(n ``%` `2` `=``=` `0``) : ` `        ``return` `(n ``/` `2``) ``*` `(n ``+` `1``) ` `  `  `   ``# If n is odd, (n+1) must be even ` `    ``else` `: ` `       ``return`  `((n ``+` `1``) ``/` `2``) ``*` `n ` `        `  `# Driver code ` `n ``=` `5` `print` `findSum(n) ` ` `  `# This code is contributed by Nikita Tiwari. `

 `// Efficient C# program to find the sum of first ` `// n natural numbers that avoid overflow if ` `// result is going to be within limits. ` `using` `System; ` ` `  `class` `GFG{ ` ` `  `    ``// Returns sum of first n natural ` `    ``// numbers ` `    ``static` `int` `findSum(``int` `n) ` `    ``{ ` `        ``if` `(n % 2 == 0) ` `            ``return` `(n / 2) * (n + 1); ` ` `  `        ``// If n is odd, (n+1) must be even ` `        ``else` `            ``return` `((n + 1) / 2) * n; ` `    ``} ` ` `  `    ``// Driver code ` `    ``public` `static` `void` `Main() ` `    ``{ ` `        ``int` `n = 5; ` `        ``Console.Write(findSum(n)); ` `    ``} ` `} ` ` `  `// This code is contributed by vt_m. `

 ` `

Output:
```15
```

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