Cube Sum of First n Natural Numbers in JavaScript
To find the cube sum, we can iterate over each natural number from 1 to βnβ. For each number, we will cube it (multiply it by itself twice) and add the result to a running total. This means we are going to add up the cubes of each number from 1 to βnβ.
There are different methods to cube the sum of the first n natural numbers in JavaScript which are as follows:
Table of Content
- Using a for-loop
- Using recursion
- Using the formula for the sum of cubes
- Using the reduce() method
Using a for-loop
The for loop is used to iterate through each natural number from 1 to βnβ inclusive, where βnβ is the parameter passed to the function. After processing all natural numbers from 1 to βnβ in the loop, the function cubeSum returns the final value of sum, which represents the total sum of the cubes of the first βnβ natural numbers.
Example: Iteratively computing the sum of cubes for numbers from 1 to N using a for loop in JavaScript.
Javascript
function cubeSum(n) { let sum = 0; for (let i = 1; i <= n; i++) { sum += i ** 3; } return sum; } console.log(cubeSum(5)); |
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Using recursion
Recursion is used to calculate the cube sum of the first βnβ natural numbers. By using recursion, the function effectively breaks down the problem of finding the cube sum of βnβ natural numbers into smaller subproblems, making the code concise and elegant.
Example: Recursively calculating the sum of cubes for numbers from 1 to N using a JavaScript function.
Javascript
function cubeSum(n) { if (n === 0) { return 0; } else { return n ** 3 + cubeSum(n - 1); } } console.log(cubeSum(4)); |
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Using the formula for the sum of cubes
A formula for the sum of cubes of the first βnβ natural numbers is used to compute the result directly. The formula is (n * (n + 1) / 2) ^ 2. This formula provides a direct and efficient way to calculate the cube sum. This approach provides a more efficient way to compute the cube sum without iterating through each natural number individually.
Example: Computing the sum of cubes for numbers from 1 to N using a closed-form mathematical formula in JavaScript
Javascript
function cubeSum(n) { return ((n * (n + 1)) / 2) ** 2; } console.log(cubeSum(6)); |
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Using the reduce() method
The reduce method is used to iterate over an array of the first n natural numbers and calculate the cube sum. We will first create an array of length n using Array.from(), and then use the reduce() method to calculate the sum of cubes. The initial value of the sum is set to 0, and for each number in the array, we add its cube to the sum. Finally, the computed sum is returned as the result.
Example: Calculating the sum of cubes for numbers from 1 to N using array generation and reduction in JavaScript.
Javascript
const cubeSum = (n) => Array.from({ length: n }, (_, i) => i + 1).reduce( (sum, num) => sum + num ** 3,0); console.log(cubeSum(7)); |
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