Sum of cubes of first n odd natural numbers

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. 
 

Output : 
 

28

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)

 

Output: 
 

496

Time Complexity: O(1)

Space Complexity: O(1)