What is the sum of first 10 natural even numbers

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 <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).
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).

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