# Project Euler Problem 6: Sum square difference

Continuing to work my way through some of of Project Euler. Problem 6 solved by my code below. Is it better to use sumOfTheSquares += i*i or utilize Math.Pow()?

The sum of the squares of the first ten natural numbers is,
$1^2 + 2^2 + ... + 10^2 = 385$

The square of the sum of the first ten natural numbers is,
$(1 + 2 + ... + 10)^2 = 55^2 = 3025$

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 one hundred natural numbers and the square of the sum.

class Program
{
static void Main(string[] args)
{
Console.WriteLine(SumSquareDifference(100));
}

static int SumSquareDifference(int upperValue)
{
int sumOfTheSquares = 0;
for (int i = 1; i <= upperValue; i++)
{
sumOfTheSquares += (int)Math.Pow(i,2); //Can't formulate this myself...
}

int squareOfTheSums =  (int)Math.Pow((upperValue + 1) * (upperValue / 2),2);

return squareOfTheSums - sumOfTheSquares;
}
}

• Just one small question - why are you not reading the overview of problem that is provided after solving almost any task on projecteuler.net (including this one)? – pgs May 12 '17 at 10:03

I don't know about the difference between sumOfTheSquares += i*i and Math.pow() but the sum of squared of first n natural numbers is as follows

$1^2 + 2^2 + 3^2 + 4^2 = \dfrac{n(n+1)(2n+1)}{6}$

So its faster than using a for loop

Edit: General way to prove this is to use mathematical induction
• Please use $\TeX$ markup for math formulae. – vnp May 12 '17 at 9:54