Sum of the Cubes of First n Natural Numbers

We will discuss here how to find the sum of the cubes of first n natural numbers.

Let us assume the required sum = S

Therefore, S = 1$^{3}$ + 2$^{3}$ + 3$^{3}$ + 4$^{3}$ + 5$^{3}$ + ................... + n$^{3}$

Now, we will use the below identity to find the value of S:

n$^{4}$ - (n - 1)$^{4}$ = 4n$^{3}$ - 6n$^{2}$ + 4n - 1

Substituting, n = 1, 2, 3, 4, 5, ............., n in the above identity, we get

                    1$^{4}$ - 0$^{4}$ = 4 ∙ 1$^{3}$ - 6 ∙ 1$^{2}$ + 4 ∙ 1 - 1

                    2$^{4}$ - 1$^{4}$ = 4 ∙ 2$^{3}$ - 6 ∙ 2$^{2}$ + 4 ∙ 2 - 1

                    3$^{4}$ - 2$^{4}$ = 4 ∙ 3$^{3}$ - 6 ∙ 3$^{2}$ + 4 ∙ 3 - 1

                    4$^{4}$ - 3$^{4}$ = 4 ∙ 4$^{3}$ - 6 ∙ 4$^{2}$ + 4 ∙ 4 - 1

                    ........ .................... ...............

             n$^{4}$ - (n - 1)$^{4}$ = 4 . n$^{3}$ - 6 ∙ n$^{2}$ + 4 ∙ n - 1

Adding we get, n$^{4}$ - 0$^{4}$ = 4(1$^{3}$ + 2$^{3}$ + 3$^{3}$ + 4$^{3}$ + ........... + n$^{3}$) - 6(1$^{2}$ + 2$^{2}$ + 3$^{2}$ + 4$^{2}$ + ........ + n$^{2}$) + 4(1 + 2 + 3 + 4 + ........ + n) - (1 + 1 + 1 + 1 + ......... n times)

⇒ n$^{4}$ = 4S - 6 ∙ \(\frac{n(n + 1)(2n + 1)}{6}\) + 4 ∙ $\frac{n(n + 1)}{2}$ - n

⇒ 4S = n$^{4}$ + n(n + 1)(2n + 1) - 2n(n + 1) + n

⇒ 4S = n$^{4}$ + n(2n$^{2}$ + 3n + 1) – 2n$^{2}$ - 2n + n

⇒ 4S = n$^{4}$ + 2n$^{3}$ + 3n$^{2}$ + n - 2n$^{2}$ - 2n + n

⇒ 4S = n$^{4}$ + 2n$^{3}$ + n$^{2}$

⇒ 4S = n$^{2}$(n$^{2}$ + 2n + 1)

⇒ 4S = n$^{2}$(n + 1)$^{2}$

Therefore, S = $\frac{n^{2}(n + 1)^{2}}{4}$ = {$\frac{n(n + 1)}{2}$}$^{2}$ = (Sum of the first n natural numbers)$^{2}$

i.e., 1$^{3}$ + 2$^{3}$ + 3$^{3}$ + 4$^{3}$ + 5$^{3}$ + ................... + n$^{3}$ = {$\frac{n(n + 1)}{2}$}$^{2}$

Thus, the sum of the cubes of first n natural numbers = {$\frac{n(n + 1)}{2}$}$^{2}$

Solved examples to find the sum of the cubes of first n natural numbers:

1. Find the sum of the cubes of first 12 natural numbers.

Solution:

Sum of the cubes of first 12 natural numbers

i.e., 1$^{3}$ + 2$^{3}$ + 3$^{3}$ + 4$^{3}$ + 5$^{3}$ + ................... + 12$^{3}$

We know the sum of the cubes of first n natural numbers (S) = {$\frac{n(n + 1)}{2}$}$^{2}$

Here n = 12

Therefore, the sum of the cubes of first 12 natural numbers = {$\frac{12(12 + 1)}{2}$}$^{2}$

= {$\frac{12 × 13}{2}$}$^{2}$

= {6 × 13}$^{2}$

= (78)$^{2}$

= 6084

2. Find the sum of the cubes of first 25 natural numbers.

Solution:

Sum of the cubes of first 25 natural numbers

i.e., 1$^{3}$ + 2$^{3}$ + 3$^{3}$ + 4$^{3}$ + 5$^{3}$ + ................... + 25$^{3}$

We know the sum of the cubes of first n natural numbers (S) = {$\frac{n(n + 1)}{2}$}$^{2}$

Here n = 25

Therefore, the sum of the cubes of first 25 natural numbers = {$\frac{25(25 + 1)}{2}$}$^{2}$

= {$\frac{12 × 26}{2}$}$^{2}$

= {25 × 13}$^{2}$

= (325)$^{2}$

= 105625

● Arithmetic Progression

• Definition of Arithmetic Progression
• General Form of an Arithmetic Progress
• Arithmetic Mean
  
• Sum of the First n Terms of an Arithmetic Progression
• Sum of the Cubes of First n Natural Numbers
  
• Sum of First n Natural Numbers
• Sum of the Squares of First n Natural Numbers
  
• Properties of Arithmetic Progression
• Selection of Terms in an Arithmetic Progression
• Arithmetic Progression Formulae
• Problems on Arithmetic Progression
• Problems on Sum of 'n' Terms of Arithmetic Progression

11 and 12 Grade Math

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