Sum of Cube of N Natural Numbers: Formula, Proof and Examples

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 [n²(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.

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 = [n²(n+1)²]/4

Where,

• S is the sum of cubes of n natural numbers
• n is the total number of natural numbers

Sum of Cubes of n Natural Numbers Proof

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

n⁴ - (n - 1)⁴ = n³ - 6n² + 4n - 1

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

• 1⁴ - 0⁴ = 1³ - 6 × 1² + 4 × 1 - 1
• 2⁴ - 1⁴ = 2³ - 6 × 2² + 4 × 2 - 1
• 3⁴ - 2⁴ = 3³ - 6 × 3² + 4×3 - 1
• .
• .
• .
• n⁴ - (n - 1)⁴ = n³ - 6n² + 4n - 1

Adding all the above expressions we get,

n⁴ - 0⁴ = 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 1² + 2² +......+ n² = [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.

Read More

• Sum of Natural Numbers
• Sum of Square of Natural Numbers
• Arithmetic Progression and Geometric Progression

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 = [n²(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 = [n²(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.