--- id: 5900f3721000cf542c50fe85 challengeType: 5 title: '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
```yml tests: - text: sumSquareDifference(10) should return 2640. testString: assert.strictEqual(sumSquareDifference(10), 2640, 'sumSquareDifference(10) should return 2640.'); - text: sumSquareDifference(20) should return 41230. testString: assert.strictEqual(sumSquareDifference(20), 41230, 'sumSquareDifference(20) should return 41230.'); - text: sumSquareDifference(100) should return 25164150. testString: assert.strictEqual(sumSquareDifference(100), 25164150, 'sumSquareDifference(100) should return 25164150.'); ```
## Challenge Seed
```js function sumSquareDifference(n) { // Good luck! return true; } sumSquareDifference(100); ```
## Solution
```js 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; } ```