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





11 and 12 Grade Math

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