Formula to find sum of the first n Natural numbers

Hello, Everyone. In this, we are going to see a proof for the formula to find sum of the first n Natural numbers.

Let S = 1 + 2 + 3 + 4 + ......... + (n-1) + n , where n is any natural number.  --------------- (1)

The r th term from the starting of the series, in the above series is r. ------- (2)

From the property of  addition of numbers, we can write S as shown below.

S = n + (n-1) + (n-2) + ......... + 2 + 1 ---------- (3) 

The r th term from the starting of the series, in the above series is n-(r-1) = n - r + 1 -------- (4)

Let us add (1) and (3) such that the corresponding terms are added as shown in the below figure.

From (2) and (4),

The r th term from the starting of the series, in the resulting series obtained from addition of (1) and (3) is equal to r + (n - r + 1) = n + 1. From this, it is obvious that the r th term of the resulting series is independent of r. Therefore, as n is a constant, every term in the resulting series is equal to n + 1.

Therefore, by adding (1) and (3), we get the following equation.

2S = (n + 1) + (n + 1) + (n + 1) + .......................... n times

Therefore, 2S = n(n + 1)

Therefore, S = Sum of the first n Natural numbers = (n(n + 1))/2.