Project Euler 6: Difference between sum of squares and square of sum I've created a (very) simple solution for Project Euler problem 6:


  Project Euler Problem 6: Sum square difference
  
  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.


Solution:


   \$25164150\$


To solve this in F# was pretty simple: I simply declared a function square which would multiply an x by itself, then a function which would run through a list of numbers and add the square of the sum of them, and subtract the sum of the squares of each.

let square x =
    x * x

let calcDiffSquareSumFromSumSquares list = 
    (list |&gt; List.sum |&gt; square) - (list |&gt; List.sumBy square)

printfn "Solution to Project Euler 6: %i" (calcDiffSquareSumFromSumSquares [ 1 .. 100 ])


On this one, I'm particularly curious if there's a functional way to run through the list only once, calculate the total sum, and calculate the squares of sums so that I wouldn't need to use two sum methods.
Looks good to me.

Couple of things to consider though:


You could split out two functions, squareOfSum and sumOfSquares. I think that would make it a little easier to read.
You could make the functions more general by using Seq.sum/Sum.sumBy instead of List.sum/List.sumBy.
While you could use fold, as joranvar mentioned, I think this would make the program much less clear and I would not recommend it (except as an exercise for learning about folds).
There is another way to attack the problem, by using the two identities


\begin{align}
1 + 2 + 3 + \cdots + n &amp;= \frac{n(n + 1)}{2}, \quad \text{and} \\
1^2 + 2^2 + 3^2 + \cdots + n^2 &amp;= \frac{n(n+1)(2n+1)}{6}.
\end{align}