## Project Euler 6

Problem 6 asks:

Find the difference between the sum of the squares of the first one hundred natural numbers and the square of the sum.

This has been the simplest problem so far, here’s my straight forward solution in Java:

public class Problem6 { public static void main(String[] args) { int sumOfSqs = 0; int sqOfSum = 0; for (int i=1; i<=100; i++) { sumOfSqs += i * i; sqOfSum += i; } sqOfSum *= sqOfSum; System.out.println(sqOfSum - sumOfSqs); } }

I decided to try and find the general case. I know from school that the sum of the natural numbers 1 to n is \(\tfrac{1}{2}n(n+1)\), so the square of the sum is \(\tfrac{1}{4}n^2(n+1)^2\). I didn’t know how to go about making a general formula for the sum of squares, so I cheated and looked on Wikipedia, I found Faulhaber’s formula which gives me the sum of squares being \(\tfrac{1}{6}n(n+1)(2n+1)\). Which implies $$\begin{align}s - t &= (\tfrac{1}{2}n(n+1))^2 - \tfrac{1}{6}n(n+1)(2n+1)\\ &= \tfrac{1}{12}n(n+1)(3n+2)(n-1)\end{align}$$ Where \(s\) is the square of the sum, and \(t\) is the sum of squares.

Now with that groundwork done it’s simple to solve this problem for very big n in a *very* fast time.