Sum of cubes of first n odd natural numbers
Last Updated :
16 Feb, 2023
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 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 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# 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
// 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.
?>
<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 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 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# 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
// 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.
?>
<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)
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