2.0 KiB
2.0 KiB
id | challengeType | title |
---|---|---|
5900f3721000cf542c50fe85 | 5 | Problem 6: Sum square difference |
Description
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;
}