4. Sum of squares of first 'n' natural numbers

The natural numbers are \(1\), \(2\), \(3\), \(4\), …

 

We need to find the value of \(1^2 + 2^2 + 3^2  + …. + n^2\).

 

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

 

Since the power of all the numbers is \(2\), put \(k = 2\) in the above identity.

 

\((x + 1)^{2 + 1} - x^{2 + 1} = (x + 1)^3 - x^3\)

 

\(= x^3 + 3x^2 + 3x + 1^3 - x^3\)

 

\(= 3x^2 + 3x + 1\)

 

\((x + 1)^3 - x^3 = 3x^2 + 3x + 1\) - - - - (I)

 

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

 

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

 

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

 

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

 

           \(\vdots\)             \(\vdots\)             \(\vdots\)

 

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

 

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

 

Add all the above equations of \(x\) values.

 

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

 

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

 

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

 

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

 

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

 

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

 

Apply sum of first \(n\) terms of natural numbers formula in \(1 + 2 + 3 + .. (n - 1) + n\).

 

 

 

 

 

 

 

 

 

Therefore, $1^2+2^2+3^2+...{(n-1)}^2+n^2=\frac{n(n+1)(2n+1)}{6}$