Sum of Sums first n natural numbers

Given a positive integer n. The task is to find the sum of the sum of first n natural number.

Examples: 

Input: n = 3
Output: 10
Explanation: 
Sum of first natural number: 1
Sum of first and second natural number: 1 + 2 = 3
Sum of first, second and third natural number = 1 + 2 + 3 = 6
Sum of sum of first three natural number = 1 + 3 + 6 = 10

Input: n = 2
Output: 4

Sum of Natural Numbers is represented by Triangular Numbers. A simple solution is to one by one add triangular numbers. 

20

Time Complexity: O(N), for traversing from 1 till N to calculate the required sum.
Auxiliary Space: O(1), as constant extra space is required.

An efficient solution is to use direct formula n(n+1)(n+2)/6
Mathematically, we need to find, Sum(((i * (i + 1))/2)), where 1 <= i <= n 
So, lets solve this summation, 

Sum = ? ((i * (i + 1))/2), where 1 <= i <= n
 = (1/2) * ? (i * (i + 1))
 = (1/2) * ? (i2 + i)
 = (1/2) * (? i2 + ? i)

We know ? i2 = n * (n + 1) * (2*n + 1) / 6 and 
? i = n * ( n + 1) / 2.
Substituting the value, we get,
Sum = (1/2) * ((n * (n + 1) * (2*n + 1) / 6) + (n * ( n + 1) / 2)) 
 = n * (n + 1)/2 [(2n + 1)/6 + 1/2]
 = n * (n + 1) * (n + 2) / 6

Below is the implementation of the above approach: 

20

Time Complexity: O(1), as constant operations are being performed.
Auxiliary Space: O(1), as constant extra space is required.