Sum of the series 1, 3, 6, 10... (Triangular Numbers)

Given n, no of elements in the series, find the summation of the series 1, 3, 6, 10....n. The series mainly represents triangular numbers.
Examples: 
 

Input: 2
Output: 4
Explanation: 1 + 3 = 4

Input: 4
Output: 20
Explanation: 1 + 3 + 6 + 10 = 20

 

A simple solution is to one by one add triangular numbers. 
 

Output: 
 

20

Time complexity : O(n)

Auxiliary Space: O(1) since using constant variables

An efficient solution is to use direct formula n(n+1)(n+2)/6
 

Let g(i) be i-th triangular number.
g(1) = 1
g(2) = 3
g(3) = 6
g(n) = n(n+1)/2

 

Let f(n) be the sum of the triangular
numbers 1 through n.
f(n) = g(1) + g(2) + ... + g(n)

Then:
f(n) = n(n+1)(n+2)/6

How can we prove this? We can prove it by induction. That is, prove two things : 
 

1. It's true for some n (n = 1, in this case).
2. If it's true for n, then it's true for n+1.

This allows us to conclude that it's true for all n >= 1.
 

Now 1) is easy. We know that f(1) = g(1) 
= 1. So it's true for n = 1.

Now for 2). Suppose it's true for n. 
Consider f(n+1). We have:
f(n+1) = g(1) + g(2) + ... + g(n) + g(n+1) 
 = f(n) + g(n+1)

Using our assumption f(n) = n(n+1)(n+2)/6 
and g(n+1) = (n+1)(n+2)/2, we have:
f(n+1) = n(n+1)(n+2)/6 + (n+1)(n+2)/2
 = n(n+1)(n+2)/6 + 3(n+1)(n+2)/6
 = (n+1)(n+2)(n+3)/6
Therefore, f(n) = n(n+1)(n+2)/6

Below is the implementation of the above approach: 
 

Output: 
 

20

Time complexity : O(1)
Auxiliary Space: O(1), since no extra space has been taken.