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++ 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> |
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++ 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> |
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++ 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 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 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 |
<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>
|
Sum of first 20 Even numbers is: 420
Time Complexity: O(1).
Auxiliary Space: O(1) since using constant variables