# Kotlin – Sum of Cubes of First N Natural Numbers

## Kotlin – Find Sum of Cubes of First N Natural Numbers

In Kotlin, you can calculate the sum of the cubes of the first N natural numbers using a loop and a simple formula. This tutorial will guide you through the process of writing a Kotlin program to find the sum of the cubes of the first N natural numbers.

## Example

### Step 1: Define the Value of N

Start by defining the value of N, which represents the number of natural numbers for which you want to calculate the sum of cubes.

``val n = 5 // Change the value of n as needed``

### Step 2: Calculate the Sum of Cubes

Create a function to calculate the sum of cubes of the first N natural numbers using a loop and the formula `sum += i * i * i`.

``````fun calculateSumOfCubes(n: Int): Int {
var sum = 0
for (i in 1..n) {
sum += i * i * i
}
return sum
}``````

### Step 3: Print the Result

Call the `calculateSumOfCubes` function with the value of N as the argument and print the result.

``````val sumOfCubes = calculateSumOfCubes(n)
println("Sum of cubes of the first \$n natural numbers is: \$sumOfCubes")``````

### Complete Kotlin Program

The following is complete Kotlin program to find the sum of the cubes of the first N natural numbers.

Kotlin Program

``````fun main() {
val n = 5 // Change the value of n as needed
val sumOfCubes = calculateSumOfCubes(n)
println("Sum of cubes of the first \$n natural numbers is: \$sumOfCubes")
}

fun calculateSumOfCubes(n: Int): Int {
var sum = 0
for (i in 1..n) {
sum += i * i * i
}
return sum
}``````

Output

``Sum of cubes of the first 5 natural numbers is: 225``

## Summary

By following these steps and using a loop to calculate the sum of cubes, you can easily find the sum of the cubes of the first N natural numbers in Kotlin.