2. Sum of first 'n' natural numbers

The natural numbers are $1$, $2$, $3$, $4$, …

 

We need to find the value of $1 + 2 + 3 + 4 + …. + n$.

 

Consider the identity $(x + 1)^{k + 1} - x^{k + 1}$.

 

Since the sum of all the numbers is $1$, put $k = 1$ in the above identity.

 

$(x + 1)^{1 + 1} - x^{1 + 1} = (x + 1)^2 - x^2$

 

$= x^2 + 2x + 1 - x^2$

 

$= 2x + 1$

 

$(x + 1)^2 - x^2 = 2x + 1$ - - - - (I)

 

Now, substitute $x = 1, 2, 3, … n$ in equation (I).

 

When $x = 1$, $2^2 - 1^2 = 2(1) + 1$

 

When $x = 2$, $3^2 - 2^2 = 2(2) + 1$

 

When $x = 3$, $4^2 - 3^2 = 2(3) + 1$

 

           $\vdots$             $\vdots$             $\vdots$

 

When $x = n - 1$, $n^2 - (n - 1)^2 = 2(n - 1) + 1$

 

When $x = n$, $(n + 1)^2 - n^2 = 2(n) + 1$

 

Add all the above equations of $x$ values.

 

$2^2 - 1^2 + 3^2 - 2^2 + 4^2 - 3^2  + \cdots + n^2 - (n - 1)^2 + (n + 1)^2 - n^2 = 2(1) + 1 + 2(2) + 1 + 2(3) + 1 + \cdots + 2(n - 1) + 1 + 2(n) + 1$

 

$2^2 + 3^2 + 4^2 + \cdots + n^2 + (n + 1)^2 - (1^2 + 2^2 + 3^2 + \cdots + (n - 1)^2 + n^2) = 2(1) + 2(2) + 2(3) + \cdots + 2(n - 1) + 2(n) + (1 + 1 + 1 + ... n \ times )$

 

By cancelling the same terms with opposite signs on the LHS, we get:

 

$(n + 1)^2 - 1^2 = 2(1 + 2 + 3 + … + (n - 1) + n) + n$

 

$n^2 + 2n + 1^2 - 1^2 = 2(1 + 2 + 3 + … + (n - 1) + n) + n$

 

$n^2 + 2n = 2(1 + 2 + 3 + … + (n - 1) + n) + n$

 

$n^2 + 2n - n = 2(1 + 2 + 3 + … + (n - 1) + n)$

 

$n^2 + n = 2(1 + 2 + 3 + … + (n - 1) + n)$

 

 

Therefore, $1+2+3+...+\left(n-1\right)+n=\frac{n(n+1)}{2}$.