1.6 KiB
1.6 KiB
id | title | challengeType | forumTopicId |
---|---|---|---|
5900f3721000cf542c50fe85 | Problem 6: Sum square difference | 5 | 302171 |
--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.
--hints--
sumSquareDifference(10)
should return a number.
assert(typeof sumSquareDifference(10) === 'number');
sumSquareDifference(10)
should return 2640.
assert.strictEqual(sumSquareDifference(10), 2640);
sumSquareDifference(20)
should return 41230.
assert.strictEqual(sumSquareDifference(20), 41230);
sumSquareDifference(100)
should return 25164150.
assert.strictEqual(sumSquareDifference(100), 25164150);
--seed--
--seed-contents--
function sumSquareDifference(n) {
return true;
}
sumSquareDifference(100);
--solutions--
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;
}