# Project Euler Problem 6: Sum square difference

Posted on

Problem

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,
12+22+...+102=385${1}^{2}+{2}^{2}+...+{10}^{2}=385$$1^2 + 2^2 + ... + 10^2 = 385$

The square of the sum of the first ten natural numbers is,
(1+2+...+10)2=552=3025$\left(1+2+...+10{\right)}^{2}={55}^{2}=3025$$(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;
}
}


Solution

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

12+22+32+42=n(n+1)(2n+1)6${1}^{2}+{2}^{2}+{3}^{2}+{4}^{2}=\frac{n\left(n+1\right)\left(2n+1\right)}{6}$$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