Given a number n, find sum of first n odd natural numbers.
Input : 2 Output : 28 1^3 + 3^3 = 28 Input : 4 Output : 496 1^3 + 3^3 + 5^3 + 7^3 = 496
A simple solution is to traverse through n odd numbers and find the sum of cubes.
C++
// Simple C++ method to find sum of cubes of // first n odd numbers. #include <iostream> using namespace std;
int cubeSum( int n)
{ int sum = 0;
for ( int i = 0; i < n; i++)
sum += (2*i + 1)*(2*i + 1)*(2*i + 1);
return sum;
} int main()
{ cout << cubeSum(2);
return 0;
} |
Java
// Java program to perform sum of // cubes of first n odd natural numbers public class GFG
{ public static int cubesum( int n)
{
int sum = 0 ;
for ( int i = 0 ; i < n; i++)
sum += ( 2 * i + 1 ) * ( 2 * i + 1 )
* ( 2 * i + 1 );
return sum;
}
// Driver function
public static void main(String args[])
{
int a = 5 ;
System.out.println(cubesum(a));
}
} // This article is published Akansh Gupta |
Python3
# Python3 program to find sum of # cubes of first n odd numbers. def cubeSum(n):
sum = 0
for i in range ( 0 , n) :
sum + = ( 2 * i + 1 ) * ( 2 * i + 1 ) * ( 2 * i + 1 )
return sum
# Driven code print (cubeSum( 2 ))
# This code is contributed by Shariq Raza |
C#
// C# program to perform sum of // cubes of first n odd natural numbers using System;
public class GFG
{ public static int cubesum( int n)
{
int sum = 0;
for ( int i = 0; i < n; i++)
sum += (2 * i + 1) * (2 * i +1)
* (2 * i + 1);
return sum;
}
// Driver function
public static void Main()
{
int a = 5;
Console.WriteLine(cubesum(a));
}
} // This code is published vt_m |
PHP
<?php // Simple PHP method to find sum of // cubes of first n odd numbers. function cubeSum( $n )
{ $sum = 0;
for ( $i = 0; $i < $n ; $i ++)
$sum += (2 * $i + 1) *
(2 * $i + 1) *
(2 * $i + 1);
return $sum ;
} // Driver Code echo cubeSum(2);
// This code is contributed by vt_m. ?> |
Javascript
<script> // Simple javascript method to find sum of cubes of // first n odd numbers. function cubeSum( n)
{ let sum = 0;
for (let i = 0; i < n; i++)
sum += (2*i + 1)*(2*i + 1)*(2*i + 1);
return sum;
} document.write(cubeSum(2));
// This code is contributed by Rajput-Ji </script> |
Output :
28
Complexity Analysis:
Time Complexity: O(n), as we are using a single traversal in the cubeSum() function.
Space Complexity:O(1)
An efficient solution is to apply the below formula.
sum = n2(2n2 - 1) How does it work? We know that sum of cubes of first n natural numbers is = n2(n+1)2 / 4 Sum of first n even numbers is 2 * n2(n+1)2 Sum of cubes of first n odd natural numbers = Sum of cubes of first 2n natural numbers - Sum of cubes of first n even natural numbers = (2n)2(2n+1)2 / 4 - 2 * n2(n+1)2 = n2(2n+1)2 - 2 * n2(n+1)2 = n2[(2n+1)2 - 2*(n+1)2] = n2(2n2 - 1)
C++
// Efficient C++ method to find sum of cubes of // first n odd numbers. #include <iostream> using namespace std;
int cubeSum( int n)
{ return n * n * (2 * n * n - 1);
} int main()
{ cout << cubeSum(4);
return 0;
} |
Java
// Java program to perform sum of // cubes of first n odd natural numbers public class GFG
{ public static int cubesum( int n)
{
return (n) * (n) * ( 2 * n * n - 1 );
}
// Driver function
public static void main(String args[])
{
int a = 4 ;
System.out.println(cubesum(a));
}
} // This code is contributed by Akansh Gupta. |
Python3
# Python3 program to find sum of # cubes of first n odd numbers. # Function to find sum of cubes # of first n odd number def cubeSum(n):
return (n * n * ( 2 * n * n - 1 ))
# Driven code print (cubeSum( 4 ))
# This code is contributed by Shariq Raza |
C#
// C# program to perform sum of // cubes of first n odd natural numbers using System;
public class GFG
{ public static int cubesum( int n)
{
return (n) * (n) * (2 * n * n - 1);
}
// Driver function
public static void Main()
{
int a = 4;
Console.WriteLine(cubesum(a));
}
} // This code is published vt_m. |
PHP
<?php // Efficient PHP method to // find sum of cubes of // first n odd numbers. function cubeSum( $n )
{ return $n * $n * (2 * $n * $n - 1);
} // Driver Code echo cubeSum(4);
// This code is contributed by vt_m. ?> |
Javascript
<script> // javascript program to perform sum of // cubes of first n odd natural numbers function cubesum(n)
{ return (n) * (n) * (2 * n * n - 1);
} // Driver function var a = 4;
document.write(cubesum(a)); // This code is contributed by Amit Katiyar </script> |
Output:
496
Complexity Analysis:
Time Complexity: O(1)
Space Complexity: O(1)