# 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 first 4 even numbers
= (2 + 4 + 6 + 8) = 20

Input : n = 20
Output : 420
```

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

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

 ` `

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

 ` `

Output:

```Sum of first 20 Even numbers is: 420
```

Time Complexity: O(1).

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