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

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