Sum of Cubes of n Natural Numbers

There are a lot of patterns in mathematics. One such pattern of numbers is the sum of cubes of n natural numbers. The cube of a number means the third exponent of a number. For example, the cube of 3 is 3³ = 27. If we keep going with greater natural numbers, you will realize that their cubes will be very large numbers. So, how to find the sum of cubes of n natural numbers in a simple way involving less time and energy? You are going to learn a simple formula for this in this article that will ease your calculations.

Natural numbers are the counting numbers that start from 1 and goes on till infinity. To find the sum of cubes of first n natural numbers means to add the cubes of a specific number of natural numbers starting from 1 and get the answer. For example, the sum of cubes of the first 5 natural numbers can be expressed as 1³ + 2³ + 3³ + 4³ + 5³, the sum of cubes of the first 10 natural numbers can be written as 1³ + 2³ + 3³ + 4³ + 5³ + 6³ + 7³ + 8³ + 9³ + 10³, and so on.

Let us look at some examples of the sum of cubes of n natural numbers.

• Sum of cubes of the first 2 natural numbers = 1³ + 2³ = 1 + 8 = 9.
• Sum of cubes of the first 3 natural numbers = 1³ + 2³ + 3³ = 1 + 8 + 27 = 36.
• Sum of cubes of the first 4 natural numbers = 1³ + 2³ + 3³ + 4³ = 1 + 8 + 27 + 64 = 100.

If you observe, as we keep on going with more natural numbers, it is becoming difficult to calculate the sum of their cubes. Here comes the importance and the need for having a formula for the sum of cubes of n natural numbers.

The sum of cubes of n natural numbers formula is given below:

If we have a series of cubes of n natural numbers as 1³ + 2³ + 3³ + 4³ + ... + n³, the formula to find the sum is,

Sum (S) = \({\{ \dfrac{n(n+1)}{2}\} }^2\) or \(\dfrac{{n}^2 {(n+1)}^2}{4}\)

where n represents the total number of natural numbers taken starting from 1.

Now, you have learned the formula for finding the sum of cubes of n natural numbers. Let us also look at how this formula is derived in the section below.

The proof of the sum of cubes of n natural numbers is important to understand so that you won't need to just memorize the formula without understanding the logic and reasoning behind it. So, let us learn the derivation of the sum of cubes formula.

Let us assume that the sum of cubes of first n natural numbers be S_n.

⇒ S_n = 1³ + 2³ + 3³ + 4³ + ... + n³ ... (equation 1)

The identity that we will be using to derive the formula is n⁴ - (n - 1)⁴ = 4n³ - 6n² + 4n - 1. If we substitute the value of n in this identity as 1, 2, 3, 4, and so on, we will get,

1⁴ - 0⁴ = 4 × 1³ - 6 × 1² + 4 × 1 - 1

2⁴ - 1⁴ = 4 × 2³ - 6 × 2² + 4 × 2 - 1

3⁴ - 2⁴ = 4 × 3³ - 6 × 3² + 4 × 3 - 1

4⁴ - 3⁴ = 4 × 4³ - 6 × 4² + 4 × 4 - 1

...

...

n⁴ - (n - 1)⁴ = 4 × n³ - 6 × n² + 4 × n - 1

Adding all these equations, we will get, n⁴ - 0⁴ = 4(1³ + 2³ + 3³ + 4³ + ........... + n³) - 6 (1² + 2² + 3² + 4² + ........ + n²) + 4 (1 + 2 + 3 + 4 + ........ + n) - (1 + 1 + 1 + 1 + ......... n times).

Now, by using equation 1, the sum of squares formula and the sum of n natural numbers formula, we will get,

n⁴ = 4S_n - 6 × [n(n + 1)(2n + 1)/6] + 4 × [n(n+1)/2] - n

⇒ 4S_n = n⁴ + n(n + 1)(2n + 1) - 2n(n + 1) + n

⇒ 4S_n = n⁴ + n(2n² + 3n + 1) – 2n² - 2n + n

⇒ 4S_n = n⁴ + 2n³ + 3n² + n - 2n² - n

⇒ 4S_n = n⁴ + 2n³ + n²

⇒ 4S_n = n² (n² + 2n + 1)

⇒ 4S_n = n² × (n + 1)²

⇒ S_n = [n² × (n + 1)²]/4

Therefore, by substituting the value of S_n, we get, 1³ + 2³ + 3³ + 4³ + ... + n³ = [n² (n + 1)²]/4. This completes the sum of cubes of n natural numbers proof. Now let us apply the formula you have learned in this article in solving some questions.

► Also Check:

• Sum of Natural Numbers Formula
• Summation Formulas
• Sum of Squares
• Natural Numbers

What is the Sum of Cubes of n Natural Numbers?

The sum of cubes of n natural numbers means finding the sum of a series of cubes of natural numbers. It can be obtained by using a simple formula S = [n² (n + 1)²]/4, where S is the sum and n is the number of natural numbers taken.

What is the Formula of Sum of Cubes of n Natural Numbers?

The formula to find the sum of cubes of n natural numbers is S = [n² (n + 1)²]/4, where n is the count of natural numbers that we take. For example, if you want to find the sum of cubes of 7 natural numbers, you will put the value of n as 7 in the formula.

What is the Sum of Cubes of First 20 Natural Numbers?

The sum of cubes of the first 20 natural numbers is 44100. Let us look at how we got the answer.

Sum of cubes of n natural numbers = [n² (n + 1)²]/4

Sum of cubes of 20 natural numbers = [20² (20 + 1)²]/4 = (400 × 441)/4 = 44100.

What is the Sum of Cubes of First 15 Natural Numbers?

The sum of cubes of the first 15 natural numbers is 14400. Let us look at how we got the answer.

Sum of cubes of 15 natural numbers = [15² (15 + 1)²]/4 = (225 × 256)/4 = 14400.

What is the Sum of Cubes of First 8 Natural Numbers?

The sum of cubes of the first 8 natural numbers is 1296. Let us look at how we got the answer.

Sum of cubes of 8 natural numbers = [8² (8 + 1)²]/4 = (64 × 81)/4 = 1296.