Sum of first n even numbers

Last Updated : 17 Jul, 2022

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

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# 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)

Auxiliary Space: O(1)

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 <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# 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).

Space Complexity: O(1) since using constant variables

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

Java

// java implementation to find sum of
// first n even numbers

import java.io.*;
import java.util.*; 
 
class GFG
{
  // function to find sum of
  // first n even numbers    
  public static 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
  public static void main(String[] args)
  {
    int n = 20;

    System.out.println("Sum of first "+n+" Even numbers is: "+evenSum(n));
  }
}
//this code is contributed by aditya942003patil

Python3

# python3 implementation to find sum of
# first n even numbers
  
# function to find sum of
# first n even numbers
def evenSum(n) :
    
    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 (int)(n/2) * (2 + tn);
    #calculate (N/2) * (a + Tn)
  
# Driver code
if __name__ == "__main__" :
    
    n = 20;
    print("Sum of first", n ,"Even numbers is:", evenSum(n));

# this code is contributed by aditya942003patil

// 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 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
    public static void Main()
    {
        int n = 20;
        
        // Function call
        Console.Write("Sum of first "+n+" Even numbers is: "+evenSum(n));
    }
}
// This code is contributed by aditya942003patil

JavaScript

<script>
// javascript implementation to find sum of
// first n even numbers
 
// function to find sum of
// first n even numbers
function evenSum(n)
{
    var 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)
    // Array to store frequency of each character
    
}
 
    // Driver program to test above
 
    var n = 20;

    // Function call
    document.write("Sum of first "+n+" Even numbers is: "+evenSum(n));
     
    // This code is contributed by aditya942003patil.
   </script>

Output

Sum of first 20 Even numbers is: 420

Time Complexity: O(1).

Auxiliary Space: O(1) since using constant variables

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