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