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Theory:

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 power 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.
 
\(2^2 - 1^2 + 3^2 - 2^2 + 4^2 - 3^2  + \cdots + n^2 - (n - 1)^2 + (n + 1)^2 - n^2 = 2(1) + 1 + 2(2) + 1 + 2(3) + 1 + \cdots + 2(n - 1) + 1 + 2(n) + 1\)
 
\(2^2 + 3^2 + 4^2 + \cdots + n^2 + (n + 1)^2 - (1^2 + 2^2 + 3^2 + \cdots + (n - 1)^2 + n^2) = 2(1) + 2(2) + 2(3) + \cdots + 2(n - 1) + 2(n) + (1 + 1 + 1 + ... n \ times )\)
 
Cancelling the same terms with opposite signs on LHS, we get:
 
\((n + 1)^2 - 1^2 = 2(1 + 2 + 3 + … + (n - 1) + n) + n\)
 
\(n^2 + 2n + 1^2 - 1^2 = 2(1 + 2 + 3 + … + (n - 1) + n) + n\)
 
\(n^2 + 2n = 2(1 + 2 + 3 + … + (n - 1) + n) + n\)
 
\(n^2 + 2n - n = 2(1 + 2 + 3 + … + (n - 1) + n)\)
 
\(n^2 + n = 2(1 + 2 + 3 + … + (n - 1) + n)\)
 
n2+n2=1+2+3+...+n1+n
 
Therefore, 1+2+3+...+n1+n=n(n+1)2.
Sum of first \(n\) natural numbers \(=\) n(n+1)2
Important!
1. The sum of first \(n\) natural numbers are also called Triangular Numbers because they form triangle shapes.
 
2. The sum of squares of first \(n\) natural numbers are also called Square Pyramidal Numbers because they form pyramid shapes with square base.