# Program to find sum of first n natural numbers

• Difficulty Level : Basic
• Last Updated : 08 Apr, 2021

Given a number n, find the sum of first natural numbers. Attention reader! Don’t stop learning now. Get hold of all the important mathematical concepts for competitive programming with the Essential Maths for CP Course at a student-friendly price. To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

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```

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 ```

## C++

 `// 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

 `// 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

 `# 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#

 `// 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.`

## PHP

 ``

## Javascript

 ``

Output :

`15`

Time Complexity: O(n)

Auxiliary Space: O(1)
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```

## C++

 `// 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;``}`

## Java

 `// 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.`

## Python

 `# 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.`

## C#

 `// 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.`

## php

 ``

## Javascript

 ``

Output:

`15`

Time Complexity: O(1)

Auxiliary Space: O(1)
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.

## C++

 `// 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;``}`

## Java

 `// 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.`

## Python

 `# 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.`

## C#

 `// 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.`

## PHP

 ``

## Javascript

 ``

Output:

`15`

Time Complexity: O(1)

Auxiliary Space: O(1)
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