Sum of Cube of N Natural Numbers: Formula, Proof and Examples (original) (raw)

Last Updated : 23 Jul, 2025

Sum of cube of n natural numbers is a mathematical pattern on which various questions were asked in competitive exam. So, the sum of cube of n natural numbers is obtained by the formula **[n 2 (n+1) 2 ]/4 where S is sum and n is number of natural numbers. Natural Numbers are the numbers started from **1 and it ends on **infinity ∞.

We have covered the Sum of cube of n natural number formula, proof and examples below.

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Sum of Cubes of n Natural Numbers

Sum of cubes of n natural numbers means adding series of the cubes of natural numbers. The natural numbers start from the 1 so the sum of cubes of first n natural numbers start from 13 + 23 +......+ n3. The sum of cubes of n natural numbers can be given by product of n square and n+1 square and divide the resultant value by 4. In other words, the sum of cubes of n natural numbers can also be calculated by squaring the sum of n natural numbers.

Sum of Cubes of n Natural Numbers Formula

The formula of sum of cubes of n natural numbers is obtained by multiplying the total number of natural numbers with the total number of natural numbers + 1 and the divide the result by 2 and then, then after the square of the obtained value. The resultant value gives the sum of cubes of n natural numbers. Below is the formula for the n natural numbers.

**S = [{n(n+1)}/2] 2 **= [n 2 (n+1) 2 ]/4

Where,

Sum of Cubes of n Natural Numbers Proof

Below is the proof of sum of cubes of first n natural numbers (consider S).

**n 4 - (n - 1) 4 = n 3 - 6n 2 + 4n - 1

Now, put n = 1, 2, 3 .... n in the above identity

Adding all the above expressions we get,

**n 4 - **0 4= 4(13 + 23 +......+ n3) - 6(12 + 22 +......+ n2) + 4(13 + 23 +......+ n3) + (1 + 1 + 1... up to n)

We know that,

**Sum of first n natural numbers 1 + 2 +......+ n = [n(n+1)] / 2

**Sum of squares of first n natural numbers 1 2 + 2 2 +......+ n 2 = [n (n+1) (2n+1)]/6

n4 - 04 = 4(13 + 23 +......+ n3) - 6([n (n+1) (2n+1)]/6) + 4([n(n+1)] / 2) + n

As we want to calculate, 13 + 23 +......+ n3 , let's represent that as S.

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

⇒ 4S = n4 + [n (n+1) (2n+1)] - [2n(n+1)] + n

⇒ 4S = n4 + [(n2 + n) (2n +1)] - 2n2 - 2n + n

⇒ 4S = n4 + 2n3 + n2+ 2n2 + n - 2n2 - 2n + n

⇒ 4S = n4 + 2n3 + n2

⇒ 4S = n2 (n2+ 2n + 1)

⇒ 4S = n2 (n + 1)2

⇒ S = [n2 (n + 1)2]/4

⇒ 13 + 23 +......+ n3 = [n2 (n + 1)2]/4 [Proved]

LHS = RHS

Alternate Method: Using Induction

To prove the formula for the sum of the cubes of the first n natural numbers, we'll use mathematical induction.

**Step 1: Base Case

Let's start with the base case, n = 1.

When n = 1,

13 = 1 [Which is True.]

**Step 2: Inductive Hypothesis

Assume that the formula holds for some arbitrary positive integer k, i.e.,

1^3 + 2^3 + 3^3 + \ldots + k^3 = \left(\frac{k(k+1)}{2}\right)^2

**Step 3: Inductive Step

We need to prove that the formula holds for k+1, assuming it holds for k.

So, we add (k+1)3 to both sides of our assumption:

1^3 + 2^3 + 3^3 + \ldots + k^3 + (k+1)^3 = \left(\frac{k(k+1)}{2}\right)^2 + (k+1)^3

L.H.S. = \left(\frac{k(k+1)}{2}\right)^2 + (k+1)^3 = \frac{k^2(k+1)^2}{4} + (k+1)^3

⇒ L.H.S. = (k+1)^2 \left(\frac{k^2}{4} + (k+1)\right)

⇒ L.H.S. = (k+1)^2 \left(\frac{k^2 + 4k + 4}{4}\right)

⇒ L.H.S. = (k+1)^2 \left(\frac{(k+2)^2}{4}\right)

⇒ L.H.S. = \left(\frac{(k+1)(k+2)}{2}\right)^2

This matches the formula for n = k+1.

Thus, it is true for all n ≥ 1.

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Sum of Cubes of n Natural Numbers Examples

**Example 1. Find the sum of cubes of first 12 natural numbers.

**Solution:

The sum of cubes of first n natural numbers is given by:

**S = [{n(n+1)}/2] 2 **= [n 2 (n+1) 2 ]/4

⇒ Sum of cubes of first 12 natural numbers = [122(12+1)2]/4

⇒ Sum of cubes of first 12 natural numbers = [122 × 132]/4

⇒ Sum of cubes of first 12 natural numbers = [144 × 169]/4

**Thus, Sum of cubes of first 12 natural numbers = 6084

**Example 2. Determine the sum of cubes of series starting from 5 and ends with 10.

**Solution:

The sum of cubes of first n natural numbers is given by:

**S = [{n(n+1)}/2] 2 **= [n 2 (n+1) 2 ]/4

To find the sum of cubes of series starting from 5 and ends with 10 we first find the sum of cubes of series 1 to 10 and the subtract the sum of cubes of 1 to 4 to get the sum of given series i.e.,

Sum of cubes of series 5 to 10 i.e., 53 +.... + 103 = Sum of cubes of series 1 to 10 i.e., 13 + 23 +.... + 103 - Sum of cubes of series 1 to 4 i.e., 13 + 23 +.... + 43

⇒ Sum of cubes of series 1 to 10 i.e., 13 + 23 +.... + 103 = [102(10+1)2]/4 = 3025

⇒ Sum of cubes of series 1 to 4 i.e., 13 + 23 +.... + 43 = [42(4+1)2]/4 = 100

Thus, Sum of cubes of series 5 to 10 i.e., 53 +.... + 103 = 3025 - 100 = 2925

Practice Questions on Sum of Cube of N Natural Numbers

Q1. Find the sum of cubes of first 20 natural numbers.

Q2. Determine the sum of cubes of series starting from 3 and ends with 9.