# Sum of first n even numbers

Given a number **n**. The problem is to find the sum of first **n** even numbers.**Examples:**

Input : n = 4 Output : 20 Sum of first4even numbers = (2 + 4 + 6 + 8) = 20 Input : n = 20 Output : 420

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**Naive Approach:** Iterate through the first **n** even numbers and add them.

## C++

`// C++ implementation to find sum of` `// first n even numbers` `#include <bits/stdc++.h>` `using` `namespace` `std;` `// function to find sum of` `// first n even numbers` `int` `evenSum(` `int` `n)` `{` ` ` `int` `curr = 2, sum = 0;` ` ` `// sum of first n even numbers` ` ` `for` `(` `int` `i = 1; i <= n; i++) {` ` ` `sum += curr;` ` ` `// next even number` ` ` `curr += 2;` ` ` `}` ` ` `// required sum` ` ` `return` `sum;` `}` `// Driver program to test above` `int` `main()` `{` ` ` `int` `n = 20;` ` ` `cout << ` `"Sum of first "` `<< n` ` ` `<< ` `" Even numbers is: "` `<< evenSum(n);` ` ` `return` `0;` `}` |

## Java

`// Java implementation to find sum` `// of first n even numbers` `import` `java.util.*;` `import` `java.lang.*;` `public` `class` `GfG{` ` ` ` ` `// function to find sum of` ` ` `// first n even numbers` ` ` `static` `int` `evenSum(` `int` `n)` ` ` `{` ` ` `int` `curr = ` `2` `, sum = ` `0` `;` ` ` `// sum of first n even numbers` ` ` `for` `(` `int` `i = ` `1` `; i <= n; i++) {` ` ` `sum += curr;` ` ` `// next even number` ` ` `curr += ` `2` `;` ` ` `}` ` ` `// required sum` ` ` `return` `sum; ` ` ` `}` ` ` ` ` `// driver function` ` ` `public` `static` `void` `main(String argc[])` ` ` `{` ` ` `int` `n = ` `20` `;` ` ` `System.out.println(` `"Sum of first "` `+ n +` ` ` `" Even numbers is: "` `+` ` ` `evenSum(n));` ` ` `}` `}` `// This code is contributed by Prerna Saini` |

## Python3

`# Python3 implementation to find sum of` `# first n even numbers` ` ` `# function to find sum of` `# first n even numbers` `def` `evensum(n):` ` ` `curr ` `=` `2` ` ` `sum` `=` `0` ` ` `i ` `=` `1` ` ` ` ` `# sum of first n even numbers` ` ` `while` `i <` `=` `n:` ` ` `sum` `+` `=` `curr` ` ` ` ` `# next even number` ` ` `curr ` `+` `=` `2` ` ` `i ` `=` `i ` `+` `1` ` ` `return` `sum` `# Driver Code` `n ` `=` `20` `print` `(` `"sum of first "` `, n, ` `"even number is: "` `,` ` ` `evensum(n))` `# This article is contributed by rishabh_jain` |

## C#

`// C# implementation to find sum` `// of first n even numbers` `using` `System;` `public` `class` `GfG {` ` ` `// function to find sum of` ` ` `// first n even numbers` ` ` `static` `int` `evenSum(` `int` `n)` ` ` `{` ` ` `int` `curr = 2, sum = 0;` ` ` `// sum of first n even numbers` ` ` `for` `(` `int` `i = 1; i <= n; i++) {` ` ` `sum += curr;` ` ` `// next even number` ` ` `curr += 2;` ` ` `}` ` ` `// required sum` ` ` `return` `sum;` ` ` `}` ` ` `// driver function` ` ` `public` `static` `void` `Main()` ` ` `{` ` ` `int` `n = 20;` ` ` ` ` `Console.WriteLine(` `"Sum of first "` `+ n` ` ` `+ ` `" Even numbers is: "` `+ evenSum(n));` ` ` `}` `}` `// This code is contributed by vt-m.` |

## PHP

`<?php` `// PHP implementation to find sum of` `// first n even numbers` `// function to find sum of` `// first n even numbers` `function` `evenSum(` `$n` `)` `{` ` ` `$curr` `= 2;` ` ` `$sum` `= 0;` ` ` `// sum of first n even numbers` ` ` `for` `(` `$i` `= 1; ` `$i` `<= ` `$n` `; ` `$i` `++) {` ` ` `$sum` `+= ` `$curr` `;` ` ` `// next even number` ` ` `$curr` `+= 2;` ` ` `}` ` ` `// required sum` ` ` `return` `$sum` `;` `}` `// Driver program to test above` ` ` `$n` `= 20;` ` ` `echo` `"Sum of first "` `.` `$n` `.` `" Even numbers is: "` `.evenSum(` `$n` `);` ` ` `// this code is contributed by mits` `?>` |

## Javascript

`<script>` `// JavaScript implementation to find sum of` `// first n even numbers` ` ` `// function to find sum of` ` ` `// first n even numbers` ` ` `function` `evenSum(n)` ` ` `{` ` ` `let curr = 2, sum = 0;` ` ` ` ` `// sum of first n even numbers` ` ` `for` `(let i = 1; i <= n; i++) {` ` ` `sum += curr;` ` ` ` ` `// next even number` ` ` `curr += 2;` ` ` `}` ` ` ` ` `// required sum` ` ` `return` `sum;` ` ` `}` ` ` ` ` `// Driver program to test above` ` ` ` ` `let n = 20;` ` ` `document.write(` `"Sum of first "` `+ n +` ` ` `" Even numbers is: "` `+ evenSum(n));` ` ` `//This code is contributed by Surbhi Tyagi` `</script>` |

**Output**

Sum of first 20 Even numbers is: 420

**Time Complexity:** O(n).**Efficient Approach:** By applying the formula given below.

Sum of first n even numbers = n * (n + 1).

**Proof:**

Sum of firstnterms of an A.P.(Arithmetic Progression) =(n/2) * [2*a + (n-1)*d].....(i) where,ais the first term of the series anddis the difference between the adjacent terms of the series. Here, a = 2, d = 2, applying these values to eq.(i), we get Sum = (n/2) * [2*2 + (n-1)*2] = (n/2) * [4 + 2*n - 2] = (n/2) * (2*n + 2) =n * (n + 1)

## C++

`// C++ implementation to find sum of` `// first n even numbers` `#include <bits/stdc++.h>` `using` `namespace` `std;` `// function to find sum of` `// first n even numbers` `int` `evenSum(` `int` `n)` `{` ` ` `// required sum` ` ` `return` `(n * (n + 1));` `}` `// Driver program to test above` `int` `main()` `{` ` ` `int` `n = 20;` ` ` `cout << ` `"Sum of first "` `<< n` ` ` `<< ` `" Even numbers is: "` `<< evenSum(n);` ` ` `return` `0;` `}` |

## Java

`// Java implementation to find sum` `// of first n even numbers` `import` `java.util.*;` `import` `java.lang.*;` `public` `class` `GfG{` ` ` ` ` `// function to find sum of` ` ` `// first n even numbers` ` ` `static` `int` `evenSum(` `int` `n)` ` ` `{` ` ` `// required sum` ` ` `return` `(n * (n + ` `1` `));` ` ` `}` ` ` ` ` `// driver function` ` ` `public` `static` `void` `main(String argc[])` ` ` `{` ` ` `int` `n = ` `20` `;` ` ` `System.out.println(` `"Sum of first "` `+ n +` ` ` `" Even numbers is: "` `+` ` ` `evenSum(n));` ` ` `}` `}` `// This code is contributed by Prerna Saini` |

## Python3

`# Python3 implementation to find` `# sum of first n even numbers` ` ` `# function to find sum of` `# first n even numbers` `def` `evensum(n):` ` ` `return` `n ` `*` `(n ` `+` `1` `)` `# Driver Code` `n ` `=` `20` `print` `(` `"sum of first"` `, n, ` `"even number is: "` `,` ` ` `evensum(n))` `# This article is contributed by rishabh_jain` |

## C#

`// C# implementation to find sum` `// of first n even numbers'` `using` `System;` `public` `class` `GfG {` ` ` `// function to find sum of` ` ` `// first n even numbers` ` ` `static` `int` `evenSum(` `int` `n)` ` ` `{` ` ` ` ` `// required sum` ` ` `return` `(n * (n + 1));` ` ` `}` ` ` `// driver function` ` ` `public` `static` `void` `Main()` ` ` `{` ` ` `int` `n = 20;` ` ` ` ` `Console.WriteLine(` `"Sum of first "` `+ n` ` ` `+ ` `" Even numbers is: "` `+ evenSum(n));` ` ` `}` `}` `// This code is contributed by vt_m` |

## PHP

`<?php` `// PHP implementation` `// to find sum of` `// first n even numbers` `// function to find sum of` `// first n even numbers` `function` `evenSum(` `$n` `)` `{` ` ` `// required sum` ` ` `return` `(` `$n` `* (` `$n` `+ 1));` `}` `// Driver Code` `$n` `= 20;` `echo` `"Sum of first "` `, ` `$n` `,` ` ` `" Even numbers is: "` `,` ` ` `evenSum(` `$n` `);` `// This code is contributed` `// by akt_mit` `?>` |

## Javascript

`<script>` `// Javascript implementation` `// to find sum of` `// first n even numbers` `// function to find sum of` `// first n even numbers` `function` `evenSum(n)` `{` ` ` `// required sum` ` ` `return` `(n * (n + 1));` `}` `// Driver Code` `let n = 20;` `document.write(` `"Sum of first "` `+ n +` ` ` `" Even numbers is: "` `,` ` ` `evenSum(n));` `// This code is contributed` `// by gfgking` `</script>` |

**Output**

Sum of first 20 Even numbers is: 420

**Time Complexity: **O(1).

**Another method:**

In this method, we have to calculate the Nth term,

The formula for finding Nth term ,**Tn = a+(n-1)d,** here, a= first term, d= common difference, n= number of term

And then we have to apply the formula for finding the sum,

the formula is,** Sn=(N/2) * (a + Tn),** here a= first term, Tn= last term, n= number of term

This formula also can be applied for the sum of odd numbers, but the series must have a same common difference.

## C++

`// C++ implementation to find sum of` `// first n even numbers` `#include <bits/stdc++.h>` `using` `namespace` `std;` `// function to find sum of` `// first n even numbers` `int` `evenSum(` `int` `n)` `{` ` ` `int` `tn = 2+(n-1)*2;` ` ` `//find Nth Term` ` ` `//calculate a+(n-1)d` ` ` `//first term is = 2` ` ` `//common difference is 2` ` ` `//first term and common difference is same all time` ` ` ` ` `// required sum` ` ` `return` `(n/2) * (2 + tn);` ` ` `//calculate (N/2) * (a + Tn)` `}` `// Driver program to test above` `int` `main()` `{` ` ` `int` `n = 20;` ` ` `cout << ` `"Sum of first "` `<< n` ` ` `<< ` `" Even numbers is: "` `<< evenSum(n);` ` ` `return` `0;` `}` `//Contributed by SoumikMondal` |

**Output**

Sum of first 20 Even numbers is: 420

**Time Complexity: **O(1).