Sum of first n even numbers

22 Jun, 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++
Java
Python3
C#
PHP
Javascript
// 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 numbersint 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 aboveint main(){    int n = 20;    cout << "Sum of first " << n         << " Even numbers is: " << evenSum(n);    return 0;}
// Java implementation to find sum// of first n even numbersimport 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 implementation to find sum of# first n even numbers     # function to find sum of# first n even numbersdef 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 Coden = 20print("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 numbersusing 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 implementation to find sum of// first n even numbers // function to find sum of// first n even numbersfunction 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?>
<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>
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++
Java
Python3
C#
PHP
Javascript
// 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 numbersint evenSum(int n){    // required sum    return (n * (n + 1));} // Driver program to test aboveint main(){    int n = 20;    cout << "Sum of first " << n         << " Even numbers is: " << evenSum(n);    return 0;}
// Java implementation to find sum// of first n even numbersimport 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 implementation to find# sum of first n even numbers     # function to find sum of# first n even numbersdef evensum(n):    return n * (n + 1) # Driver Coden = 20print("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 implementation// to find sum of// first n even numbers // function to find sum of// first n even numbersfunction 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?>
<script>// Javascript implementation// to find sum of// first n even numbers // function to find sum of// first n even numbersfunction evenSum(n){     // required sum    return (n * (n + 1));} // Driver Codelet n = 20;document.write("Sum of first " + n +    " Even numbers is: " ,              evenSum(n)); // This code is contributed// by gfgking</script>
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 numbersint 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 aboveint main(){    int n = 20;    cout << "Sum of first " << n         << " Even numbers is: " << evenSum(n);    return 0;}//Contributed by SoumikMondal
Sum of first 20 Even numbers is: 420
Time Complexity: O(1).
Space Complexity: O(1) since using constant variables 
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Mithun Kumar
surbhityagi15
gfgking
SoumikMondal
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