Sum of Cubes

## Sum of Cubes of n Natural Numbers

Enter the value of n:

### The sum of cubes of the first n natural numbers can be expressed as:

1³ + 2³ + 3³ + … + n³

This is a common mathematical problem that arises in various fields such as calculus, algebra, and even physics. In this post, we will explore different methods to calculate the sum of cubes of n natural numbers.

### Method 1: Using Formula

There is a formula to calculate the sum of cubes of the first n natural numbers:

1³ + 2³ + 3³ + … + n³ = [n(n+1)/2]²

Using this formula, we can directly calculate the sum of cubes for any value of n. Let’s take an example to understand this method.

Example: Find the sum of cubes of the first 5 natural numbers.

Solution: Using the formula, we have: 1³ + 2³ + 3³ + 4³ + 5³ = [5(5+1)/2]² = (5×6/2)² = 15² = 225

Therefore, the sum of cubes of the first 5 natural numbers is 225.

### Method 2: Using Iteration

Another method to calculate the sum of cubes is by using iteration. We can write a loop to iterate through the numbers and add the cube of each number to a variable that stores the sum.

Example: Find the sum of cubes of the first 4 natural numbers using iteration.

Solution: Let’s use a for loop to iterate through the numbers and add their cubes to a sum variable:

sum = 0 for i in range(1, 5): sum += i**3

print(“The sum of cubes of the first 4 natural numbers is:”, sum)

Output: The sum of cubes of the first 4 natural numbers is: 100

Therefore, the sum of cubes of the first 4 natural numbers is 100.

### Method 3: Using Recursive Formula

We can also use a recursive formula to calculate the sum of cubes. The formula is:

1³ + 2³ + 3³ + … + n³ = n³ + (n-1)³ + (n-2)³ + … + 1³

Using this formula, we can recursively calculate the sum of cubes of the first n natural numbers.

Example: Find the sum of cubes of the first 6 natural numbers using the recursive formula.

Solution: Using the recursive formula, we have:

1³ + 2³ + 3³ + 4³ + 5³ + 6³ = 6³ + 5³ + 4³ + 3³ + 2³ + 1³

= 216 + 125 + 64 + 27 + 8 + 1

= 441

Therefore, the sum of cubes of the first 6 natural numbers is 441.

Here are a few more examples:

Example 1: Find the sum of cubes of the first 4 natural numbers. Solution: The first 4 natural numbers are 1, 2, 3, and 4. Therefore, the sum of cubes of these numbers can be found as: 1³ + 2³ + 3³ + 4³ = 1 + 8 + 27 + 64 = 100.

Example 2: Find the sum of cubes of the first 5 even numbers. Solution: The first 5 even numbers are 2, 4, 6, 8, and 10. Therefore, the sum of cubes of these numbers can be found as: 2³ + 4³ + 6³ + 8³ + 10³ = 8 + 64 + 216 + 512 + 1000 = 1800.

Example 3: Find the sum of cubes of the first 6 odd numbers. Solution: The first 6 odd numbers are 1, 3, 5, 7, 9, and 11. Therefore, the sum of cubes of these numbers can be found as: 1³ + 3³ + 5³ + 7³ + 9³ + 11³ = 1 + 27 + 125 + 343 + 729 + 1331 = 2456.

Example 4: Find the sum of cubes of the first 8 prime numbers. Solution: The first 8 prime numbers are 2, 3, 5, 7, 11, 13, 17, and 19. Therefore, the sum of cubes of these numbers can be found as: 2³ + 3³ + 5³ + 7³ + 11³ + 13³ + 17³ + 19³ = 8 + 27 + 125 + 343 + 1331 + 2197 + 4913 + 6859 = 15603.

Example 5: Find the sum of cubes of the first 10 Fibonacci numbers. Solution: The first 10 Fibonacci numbers are 1, 1, 2, 3, 5, 8, 13, 21, 34, and 55. Therefore, the sum of cubes of these numbers can be found as: 1³ + 1³ + 2³ + 3³ + 5³ + 8³ + 13³ + 21³ + 34³ + 55³ = 1 + 1 + 8 + 27 + 125 + 512 + 2197 + 9261 + 39304 + 166375 = 217611.

The sum of cubes of n natural numbers is a commonly used mathematical concept. In this post, we discussed different methods to calculate the sum of cubes, including using a formula, iteration, and a recursive formula. Each method has its own advantages and disadvantages, and depending on the situation, one method may be more suitable than the others.