Probability for three randomly chosen numbers to be in AP

Given a number n and an array containing 1 to (2n+1) consecutive numbers. Three elements are chosen at random. Find the probability that the elements chosen are in A.P.

Examples:  

Input : n = 2
Output : 0.4
Explanation:
The array would be {1, 2, 3, 4, 5}
Out of all elements, triplets which are in AP: {1, 2, 3}, {2, 3, 4}, {3, 4, 5}, {1, 3, 5}
No of ways to choose elements from the array: 10 (⁵C₃) 
So, probability = 4/10 = 0.4

Input : n = 5
Output : 0.1515

The number of ways to select any 3 numbers from (2n+1) numbers are:^{(2n + 1)} C ₃ 
Now, for the numbers to be in AP: 
with common difference 1---{1, 2, 3}, {2, 3, 4}, {3, 4, 5}...{2n-1, 2n, 2n+1} 
with common difference 2---{1, 3, 5}, {2, 4, 6}, {3, 5, 7}...{2n-3, 2n-1, 2n+1} 
with common difference n--- {1, n+1, 2n+1} 
Therefore, Total number of AP group of 3 numbers in (2n+1) numbers are: 
(2n - 1)+(2n - 3)+(2n - 5) +...+ 3 + 1 = n * n (Sum of first n odd numbers is n * n ) 
So, probability for 3 randomly chosen numbers in (2n + 1) consecutive numbers to be in AP = (n * n) / ^{(2n + 1)} C ₃ = 3 n / (4 (n * n) - 1)
 

Below is the implementation of the above approach:

0.151515

Time Complexity: O(1)
Auxiliary Space: O(1)