Sum of Squares of First N Natural Numbers in JavaScript
The sum of squares of the first n natural numbers is a mathematical problem of finding the sum of squares of each number up to a given positive number n. We can find the sum of squares of first n natural numbers in JavaScript using the approaches listed below.
Table of Content
- Sum of Square of First N Natural Numbers using for Loop
- Sum of Square of First N Natural Numbers using reduce() Method
- Sum of Square of First N Natural Numbers using Mathematical Formula
- Sum of Square of First N Natural Numbers using Recursion
Sum of Square of First N Natural Numbers using for Loop
The for loop is used to calculate the sum of the squares of first N natural numbers. First, initialize a variable sum with 0 then iterating over the numbers from 1 to N and adding the square of each number to the sum variable.
Syntax:
for (initialization; condition; update) {
// Code
}
Example: This example uses the for loop to get a sum of squares of first n natural numbers in JavaScript.
let n = 8;
let sum = 0;
for (let i = 1; i <= n; i++) {
sum += i * i;
}
console.log(sum);
Output
204
Sum of Square of First N Natural Numbers using reduce() Method
Creates an array of numbers from 1 to n, then use the reduce() method to accumulate the sum of squares, by starting with an initial value of 0.
Syntax:
array.reduce(callback(accumulator, currentValue[, index[, array]])[, initialValue])
Example: This example uses the reduce() method to get a sum of squares of first n natural numbers in JavaScript.
let n = 8;
let numbers = Array.from({ length: n }, (_, i) => i + 1);
let sum = numbers.reduce((acc, num) => acc + num * num, 0);
console.log(sum);
Output
204
Sum of Square of First N Natural Numbers using Mathematical Formula
This approach uses the direct mathematical formula (n * (n + 1) * (2 * n + 1)) / 6 to calculate the sum of the squares of the first n natural numbers.
Syntax:
Sum = (n * (n + 1) * (2 * n + 1)) / 6;
Example: This example uses the mathematical formula to get a sum of squares of first n natural numbers in JavaScript.
let n = 8;
let sum = (n * (n + 1) * (2 * n + 1)) / 6;
console.log(sum);
Output
204
Sum of Square of First N Natural Numbers using Recursion:
This approach uses a recursive function to find the sum of squares. The base case has the current number greater than n, in which case it returns 0; otherwise, it gets the square of the current number and adds this to the result of next number’s recursive call.
Syntax
const sum = (n, current = 1) => (current > n) ? 0 : current * current + sum(n, current + 1);
Example: This example uses recursion to get a sum of squares of first n natural numbers in JavaScript
function sumOfSquaresRecursion(n, current = 1) {
if (current > n) {
return 0;
}
return (
current * current +
sumOfSquaresRecursion(n, current + 1)
);
}
// Example usage:
let n = 8;
console.log(sumOfSquaresRecursion(n));
Output
204