As you all know, there are many patterns in mathematics. One of them is the sum of **Cubes of n Natural Numbers.**

The cube of a number means the third exponent of a number. In other words, a cube number is generated when a number has been multiplied by itself twice. The symbol for cubed is 3.

For example:-

27 is a cube number because it’s 3 x 3 x 3 (3 multiplied by itself twice); this is also written as 33 (“three cubed”) or we can say that the cube of 3 is 33 = 27.

The first 10 cube numbers are:

- 1 x 1 x 1 or 1
^{3}= 1 - 2 x 2 x 2 or 2
^{3}= 8 - 3 x 3 x 3 or 3
^{3}= 27 - 4 x 4 x 4 or 4
^{3}= 64 - 5 x 5 x 5 or 5
^{3}= 125 - 6 x 6 x 6 or 6
^{3}= 216 - 7 x 7 x 7 or 7
^{3}= 343 - 8 x 8 x 8 or 8
^{3}= 512 - 9 x 9 x 9 or 9
^{3}= 729 - 10 x 10 x 10 or 10
^{3}= 1,000

Finding the cube of smaller numbers is quite easy but when we move greater natural numbers, we will realize that their cubes will be very large numbers.

So, we have to find the sum of **Cubes of n Natural Numbers** in a simple way which involves less time and energy.

Table of Contents

**Sum of Cubes of First n Natural Numbers**

Natural numbers are the numbers that we can count and start from 1 and go on till infinity.

To find the sum of cubes of first n natural numbers means that we have to add the cubes of a specific number of natural numbers starting from 1 and we can get the answer.

For example: –

Sum of cubes of the first 5 natural numbers can be written as 1^{3} + 2^{3} + 3^{3}+ 4^{3}+ 5^{3}.

Similarly, the sum of cubes of the first 10 natural numbers can be expressed as

1^{3} + 2^{3} + 3^{3}+ 4^{3}+ 5^{3} + 6^{3 }+ 7^{3 }+8^{3 }+ 9^{3 }+10^{3 }and so on.

Now, Let us consider some examples of the sum of **Cubes of n Natural Numbers**.

1. Sum of cubes of the first 2 natural numbers is 1^{3} + 2^{3} = 1 + 8 = 9

2. Sum of cubes of the first 3 natural numbers is 1^{3} + 2^{3} + 3^{3} = 1 + 8 + 27 = 36

3. Sum of cubes of the first 4 natural numbers 1^{3} + 2^{3} + 3^{3} + 4^{3} = 1 + 8 + 27 + 64 = 100

When you carefully observe the examples, you will see that as we keep on moving with more natural numbers, it is becoming difficult to calculate the sum of their cubes.

At this point, the need and importance of having a formula to compute the sum of **Cubes of n Natural Numbers**.

## Sum of Cubes of n Natural Numbers Formula

Formula of the sum of cubes of n natural numbers is given below

1^{3} + 2^{3} + 3^{3} + 4^{3} +………+ n^{3}

Sum = [ n(n+1)/2]^{2}

or we can write Sum = n^{2}(n+1)^{2}/4

The formula states that if we have a series of **Cubes of n Natural Numbers ** as 1^{3} + 2^{3} + 3^{3} + 4^{3} +………+ n^{3}

we can find the sum by using the given formula.

Sum (S) = [ n(n+1)/2]^{2}

or n^{2}(n+1)^{2}/4

Here, n represents all natural numbers starting from 1.

Now, Let us understand how this formula is derived below.

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**Sum of Cubes of n Natural Numbers Proof**

Suppose that the sum of cubes of first n natural numbers is Sn.

S_{n} = 1^{3} + 2^{3} + 3^{3} + 4^{3} +………+ n^{3}_{…….}

Here, we are using identity n^{4 }– (n – 1)^{4} = 4n^{3 }– 6n^{2} + 4n – 1 to derive the formula .when 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 ( using n^{4} – (n – 1)^{4} = 4n^{3} – 6n^{2} + 4n – 1 )

Similarly,

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

…

…

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

Now, add 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

Now, use equation 1 which is 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} + n2

⇒ 4Sn = n2 (n2 + 2n + 1)

⇒ 4S_{n} = n^{2} × (n + 1)^{2}

⇒ S_{n} = [n^{2} × (n + 1)^{2}]/4

Hence, by substituting the value of S_{n} we will get, 1^{3} + 2^{3} + 3^{3} + 4^{3} +………+ n^{3 }_{= 13 + 23 + 33 + 43 +………+ n3 = n2 (n+1)2/4}

Therefore, the sum of **Cubes of n Natural Numbers ** is proved.

Read More : Remember Multiplication Tables of Any Number Using Mental Mathematics

Now, let’s practice some questions on this Formula: –

**1. Find the Sum of Cubes of the First 10 Natural Numbers.**

Answer:

Here, the sum of cubes of n natural numbers formula is: – S_{n} = [n^{2} (n + 1)^{2}]/4

where S is the sum.

According to the question, the value of n is 10.

Now, substituting the value of n in the formula, we will 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

Hence, the sum of cubes of the first 10 natural numbers is 3025.

**2. Find the Sum of Cubes of the First 15 Natural Numbers.**

S_{n} = [n^{2} × (n + 1)^{2}]/4

where S is the sum.

According to the question, the value of n is 15.

Now, substituting the value of n in the formula, we will get,

S_{15} = 15^{2} × (15+1)^{2}/4

S_{15 }= 15^{2} × 16^{2}/4

S_{15} = 225 × 256/4

S_{15} = 225 x 64

S_{15} = 14400

Hence, the sum of cubes of the first 15 natural numbers is 14400.

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**3. Find the Sum of Cubes of Natural Numbers from 4 to 14.**

To find the sum of cubes of numbers from 4 to 14, first we have to find the sum of cubes of the first 14 natural numbers and then find the sum of cubes of the first 3 natural numbers.

At last, we can find the sum of cubes from 4 to 14 by subtracting both the values.

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 3 natural numbers = 3^{2} × (3+1)^{2}/4

= 3^{2} × 4^{2}/4

= 9 × 4

= 36

Hence, the sum of cubes of natural numbers from 4 to 14 = 11025 – 36 = 10989.

**Practice Questions: –**

1. Find the sum of cubes of the first 18 natural numbers?

2. Find the sum of cubes of the first 7 natural numbers?

3. Find the sum of cubes of the first 35 natural numbers?

4. Find the sum of cubes of the natural numbers from 10 to 37?

5. Find the sum of cubes of the natural numbers from 7 to 28?

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