Find the sum of cubes of the first N natural numbers in C#

We are given a number N, and we need to calculate the sum of the cubes of the first N natural numbers. In this article, we are going to learn how we can find the sum of cubes of the first N natural numbers in C#.

Problem Description

The task is to compute the sum: 1³ + 2³ + 3³ + ... + N³ for a given positive integer N.

Example 1

• Input: N = 3
• Output: 36

Explanation:

The cubes of the first 3 natural numbers are
1³ = 1, 2³ = 8, 3³ = 27
The sum of these cubes is: 1 + 8 + 27 = 36

Example 2

• Input: N = 5
• Output: 225

Explanation:

The cubes of the first 5 natural numbers are
1³ = 1, 2³ = 8, 3³ = 27, 4³ = 64, 5³ = 125
The sum of these cubes is: 1 + 8 + 27 + 64 + 125 = 225

Using Iterative Approach

This is a simple and direct approach to find the sum of cubes. We use a loop to calculate the cube of each number from 1 to N and add it to a cumulative sum variable.

Steps for Implementation

1. Initialize a variable sum to 0.
2. Loop through each number from 1 to N.
3. For each number, calculate its cube and add it to the sum.
4. Return the final value of the sum.

Example

using System;

class Program {
    static int SumOfCubesIterative(int n) {
        int sum = 0;
        for (int i = 1; i <= n; i++) {
            sum += i * i * i;
        }
        return sum;
    }

    static void Main() {
        int N = 3;
        int result = SumOfCubesIterative(N);
        Console.WriteLine("The sum of cubes of the first {0} natural numbers is: {1}", N, result);
        
        N = 5;
        result = SumOfCubesIterative(N);
        Console.WriteLine("The sum of cubes of the first {0} natural numbers is: {1}", N, result);
    }
}

The output of the above code is

The sum of cubes of the first 3 natural numbers is: 36
The sum of cubes of the first 5 natural numbers is: 225

Time Complexity: O(N)
Space Complexity: O(1)

Using Formula-Based Approach

In this method, we use a mathematical formula to find the sum of cubes without iterating through numbers. The formula for the sum of cubes of the first N natural numbers is

Sum = [N × (N + 1) / 2]²

This formula is based on the mathematical identity that the sum of cubes equals the square of the sum of natural numbers.

Steps for Implementation

1. Calculate the sum of first N natural numbers using N × (N + 1) / 2.
2. Square the result to get the sum of cubes.
3. Return the final result.

Example

using System;

class Program {
    static int SumOfCubesFormula(int n) {
        // Calculate sum using the formula: [n*(n+1)/2]²
        int sumOfNumbers = n * (n + 1) / 2;
        return sumOfNumbers * sumOfNumbers;
    }

    static void Main() {
        int N = 5;
        int result = SumOfCubesFormula(N);
        Console.WriteLine("The sum of cubes of the first {0} natural numbers is: {1}", N, result);
        
        N = 4;
        result = SumOfCubesFormula(N);
        Console.WriteLine("The sum of cubes of the first {0} natural numbers is: {1}", N, result);
    }
}

The output of the above code is

The sum of cubes of the first 5 natural numbers is: 225
The sum of cubes of the first 4 natural numbers is: 100

Time Complexity: O(1)
Space Complexity: O(1)

Using Recursive Approach

In this approach, we use recursive algorithm to calculate the sum of cubes. For each recursive call, we add the cube of the current number to the result of the remaining numbers.

Steps for Implementation

1. Define a recursive function SumOfCubesRecursive.
2. Define a base case: If n = 0, return 0.
3. For recursive case return cube of n + recursive function(n - 1), i.e., n³ + SumOfCubesRecursive(n - 1).

Example

using System;

class Program {
    static int SumOfCubesRecursive(int n) {
        // Base case
        if (n == 0) return 0;
        
        // Recursive case: n³ + sum of cubes of (n-1) numbers
        return n * n * n + SumOfCubesRecursive(n - 1);
    }

    static void Main() {
        int N = 4;
        int result = SumOfCubesRecursive(N);
        Console.WriteLine("The sum of cubes of the first {0} natural numbers is: {1}", N, result);
        
        N = 3;
        result = SumOfCubesRecursive(N);
        Console.WriteLine("The sum of cubes of the first {0} natural numbers is: {1}", N, result);
    }
}

The output of the above code is

The sum of cubes of the first 4 natural numbers is: 100
The sum of cubes of the first 3 natural numbers is: 36

Time Complexity: O(N)
Space Complexity: O(N) due to recursive call stack

Comparison of Approaches

Conclusion

The sum of cubes of first N natural numbers can be calculated using three different approaches in C#. The formula-based approach [N×(N+1)/2]² is the most efficient with O(1) complexity, while iterative and recursive approaches provide easier-to-understand implementations with O(N) complexity.