# Sum of first n even numbers

• Difficulty Level : Easy
• Last Updated : 17 Aug, 2021

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

```Input : n = 4
Output : 20
Sum of first 4 even 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 ` `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

 ``

## Javascript

 ``
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 first n terms of an A.P.(Arithmetic Progression)
= (n/2) * [2*a + (n-1)*d].....(i)
where, a is the first term of the series and d is
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 ``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

 ``

## Javascript

 ``
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 ``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).

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