PUMPA - SMART LEARNING

எங்கள் ஆசிரியர்களுடன் 1-ஆன்-1 ஆலோசனை நேரத்தைப் பெறுங்கள். டாப்பர் ஆவதற்கு நாங்கள் பயிற்சி அளிப்போம்

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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:
 
\((n + 1)^2 - 1^2 = 2(1 + 2 + 3 + … + (n - 1) + n) + n\)
 
\(n^2 + 2n + 1^2 - 1^2 = 2(1 + 2 + 3 + … + (n - 1) + n) + n\)
 
\(n^2 + 2n = 2(1 + 2 + 3 + … + (n - 1) + n) + n\)
 
\(n^2 + 2n - n = 2(1 + 2 + 3 + … + (n - 1) + n)\)
 
\(n^2 + n = 2(1 + 2 + 3 + … + (n - 1) + n)\)
 
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 the first \(n\) natural numbers is also called a Triangular Number because they form triangle shapes.
 
2. The sum of squares of the first \(n\) natural numbers are also called Square Pyramidal Numbers because they form pyramid shapes with a square base.