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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 3^{3} = 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.
1.  Sum of Cubes of First n Natural Numbers 
2.  Sum of Cubes of n Natural Numbers Formula 
3.  Sum of Cubes of n Natural Numbers Proof 
4.  FAQs on Sum of Cubes of n Natural Numbers 
Sum of Cubes of First n Natural Numbers
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 1^{3} + 2^{3} + 3^{3} + 4^{3} + 5^{3}, the sum of cubes of the first 10 natural numbers can be written as 1^{3} + 2^{3} + 3^{3} + 4^{3} + 5^{3} + 6^{3} + 7^{3} + 8^{3} + 9^{3} + 10^{3}, 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 = 1^{3} + 2^{3} = 1 + 8 = 9.
 Sum of cubes of the first 3 natural numbers = 1^{3} + 2^{3} + 3^{3} = 1 + 8 + 27 = 36.
 Sum of cubes of the first 4 natural numbers = 1^{3} + 2^{3} + 3^{3} + 4^{3} = 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.
Sum of Cubes of n Natural Numbers Formula
The sum of cubes of n natural numbers formula is given below:
If we have a series of cubes of n natural numbers as 1^{3} + 2^{3} + 3^{3} + 4^{3} + ... + n^{3}, 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.
Sum of Cubes of n Natural Numbers Proof
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} = 1^{3} + 2^{3} + 3^{3} + 4^{3} + ... + n^{3} ... (equation 1)
The identity that we will be using to derive the formula is n^{4}  (n  1)^{4} = 4n^{3}  6n^{2} + 4n  1. If we substitute the value of n in this identity as 1, 2, 3, 4, and so on, we will 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 all these equations, we will 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).
Now, by using equation 1, the sum of squares formula and the sum of n natural numbers formula, we will get,
n^{4} = 4S_{n}  6 × [n(n + 1)(2n + 1)/6] + 4 × [n(n+1)/2]  n
⇒ 4S_{n} = n^{4} + n(n + 1)(2n + 1)  2n(n + 1) + n
⇒ 4S_{n} = n^{4} + n(2n^{2} + 3n + 1) – 2n^{2}  2n + n
⇒ 4S_{n} = n^{4} + 2n^{3} + 3n^{2} + n  2n^{2}  n
⇒ 4S_{n} = n^{4} + 2n^{3} + n^{2}
⇒ 4S_{n} = n^{2} (n^{2} + 2n + 1)
⇒ 4S_{n} = n^{2} × (n + 1)^{2}
⇒ S_{n} = [n^{2} × (n + 1)^{2}]/4
Therefore, by substituting the value of S_{n}, we get, 1^{3} + 2^{3} + 3^{3} + 4^{3} + ... + n^{3} = [n^{2} (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.
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Sum of Cubes of n Natural Numbers Examples

Example 1: Find the sum of cubes of the first 10 natural numbers.
Solution: By applying the sum of cubes of n natural numbers formula, we have S_{n} = [n^{2} (n + 1)^{2}]/4, where S is the required sum. In the given question, the value of n is 10. So, by substituting the value of n, we get,
S_{10} = 10^{2} × (10+1)^{2}/4
⇒ S_{10} = 10^{2} × 11^{2}/4
⇒ S_{10} = 100 × 121/4
⇒ S_{10} = 25 × 121
⇒ S_{10} = 3025
Therefore, the sum of cubes of the first 10 natural numbers is 3025.

Example 2: Find the sum of cubes of natural numbers from 5 to 14.
Solution: To find the sum of cubes of numbers from 5 to 14, we will first find the sum of cubes of the first 14 natural numbers and then we will find the sum of cubes of the first 4 natural numbers. Then we subtract the values obtained to get the answer.
Sum of cubes of n natural numbers = [n^{2} (n + 1)^{2}]/4
Sum of cubes of first 14 natural numbers = 14^{2} × (14+1)^{2}/4
= 14^{2} × 15^{2}/4
= 196 × 225/4
= 49 × 225
= 11025
Now, the sum of cubes of first 4 natural numbers = 4^{2} × (4+1)^{2}/4
= 4^{2} × 5^{2}/4
= 4 × 25
= 100
Therefore, the sum of cubes of natural numbers from 5 to 14 = 11025  100 = 10925.
FAQs on Sum of Cubes of n 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 = [n^{2} (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 = [n^{2} (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 = [n^{2} (n + 1)^{2}]/4
Sum of cubes of 20 natural numbers = [20^{2} (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 = [15^{2} (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 = [8^{2} (8 + 1)^{2}]/4 = (64 × 81)/4 = 1296.
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