freeCodeCamp/curriculum/challenges/english/08-coding-interview-prep/project-euler/problem-6-sum-square-difference.english.md

id, challengeType, title

id	challengeType	title
5900f3721000cf542c50fe85	5	Problem 6: Sum square difference

Description

The sum of the squares of the first ten natural numbers is,

1² + 2² + ... + 10² = 385

The square of the sum of the first ten natural numbers is,

(1 + 2 + ... + 10)² = 55² = 3025

Hence the difference between the sum of the squares of the first ten natural numbers and the square of the sum is 3025 − 385 = 2640. Find the difference between the sum of the squares of the first n natural numbers and the square of the sum.

Instructions

Tests

tests:
  - text: <code>sumSquareDifference(10)</code> should return 2640.
    testString: assert.strictEqual(sumSquareDifference(10), 2640, '<code>sumSquareDifference(10)</code> should return 2640.');
  - text: <code>sumSquareDifference(20)</code> should return 41230.
    testString: assert.strictEqual(sumSquareDifference(20), 41230, '<code>sumSquareDifference(20)</code> should return 41230.');
  - text: <code>sumSquareDifference(100)</code> should return 25164150.
    testString: assert.strictEqual(sumSquareDifference(100), 25164150, '<code>sumSquareDifference(100)</code> should return 25164150.');

Challenge Seed

function sumSquareDifference(n) {
  // Good luck!
  return true;
}

sumSquareDifference(100);

Solution

const sumSquareDifference = (number)=>{
  let squareOfSum = Math.pow(sumOfArithmeticSeries(1,1,number),2);
  let sumOfSquare = sumOfSquareOfNumbers(number);
 return squareOfSum - sumOfSquare;
}

function sumOfArithmeticSeries(a,d,n){
  return (n/2)*(2*a+(n-1)*d);
}

function sumOfSquareOfNumbers(n){
 return (n*(n+1)*(2*n+1))/6;
}