5. Sum of cubes of first 'n' natural numbers

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

 

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

 

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

 

Since the power of all the numbers is \(3\), put \(k = 3\) in the above identity.

 

\((x + 1)^{3 + 1} - x^{3 + 1} = (x + 1)^4 - x^4\)

 

\(= ((x + 1)^2)^2 - x^4\)

 

\(= (x^2 + 2x + 1^2)^2 - x^4\)

 

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

 

\(= 4x^3 + 6x^2 + 4x + 1\)

 

\((x + 1)^4 - x^4 = 4x^3 + 6x^2 + 4x + 1\) - - - - (I)

 

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

 

When \(x = 1\), \(2^4 - 1^4 = 4(1)^3 + 6(1)^2 + 4(1) + 1\)

 

When \(x = 2\), \(3^4 - 2^4 = 4(2)^3 + 6(2)^2 + 4(2) + 1\)

 

When \(x = 3\), \(4^4 - 3^4 = 4(3)^3 + 6(3)^2 + 4(3) + 1\)

 

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

 

When \(x = n - 1\), \(n^4 - (n - 1)^4 = 4(n - 1)^3 + 6(n - 1)^2 + 4(n - 1) + 1\)

 

When \(x = n\), \((n + 1)^4 - n^4 = 4(n)^3 + 6(n)^2 + 4(n) + 1\)

 

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

 

\(2^4 - 1^4 + 3^4 - 2^4 + 4^4 - 3^4  + \cdots + n^4 - (n - 1)^4 + (n + 1)^4 - n^4 = 4(1)^3 + 6(1)^2 + 4(1) + 1 + 4(2)^3 + 6(2)^2 + 4(2) + 1 + 4(3)^3 + 6(3)^2 + 4(3) + 1 + \cdots + 4(n - 1)^3 + 6(n - 1)^2 + 4(n - 1) + 1 + 4(n)^3 + 6(n)^2 + 4(n) + 1\)

 

\(2^4 + 3^4 + 4^4 + \cdots + n^4 + (n + 1)^4 - (1^4 + 2^4 + 3^4 + \cdots + (n - 1)^4 + n^4) = 4(1^3 + 2^3 + 3^3 + … + (n - 1)^3 + n^3) + 6(1^2 + 2^2 + 3^2 + \cdots + (n - 1)^2 + n^2) + 4(1 + 2 + 3 + \cdots (n - 1) + n) + (1 + 1 + 1 + ... n \ times )\)

 

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

 

\((n + 1)^4 - 1^4 = 4(1^3 + 2^3 + 3^3 + … + (n - 1)^3 + n^3) + 6(1^2 + 2^2 + 3^2 + … (n - 1)^2 + n^2) + 4(1 + 2 + 3 + … (n - 1) + n) + n\)

 

\(n^4 + 4n^3 + 6n^2 + 4n + 1^4 - 1^4 = 4(1^3 + 2^3 + 3^3 + … + (n - 1)^3 + n^3) + 6(1^2 + 2^2 + 3^2 + … (n - 1)^2 + n^2) + 4(1 + 2 + 3 + … (n - 1) + n) + n\)

 

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

 

And, apply sum of squares of first \(n\) terms of natural numbers formula in \(1^2 + 2^2 + 3^2 + .. (n - 1)^2 + n^2\).

 

 

 

 

 

 

 

 

 

 

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