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++ 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 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 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 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
?>




<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++ 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 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 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 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
?>




<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++ 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).




Article Tags :
Mathematical
series
Practice Tags :
Mathematical
series