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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 <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 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#




// 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).
 




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