Created
August 27, 2011 14:40
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Problem #6 from http://projecteuler.net/
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(* | |
Problem 6 | |
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 one hundred natural numbers and the square of the sum. | |
(a + b + c + d + e)^2 =a^2 + b^2 + c^2 + d^2 + e^2 + 2a(b + c + d + e) + 2b(c + d + e) + 2c(d + e) + 2de | |
*) | |
let problem_6 n = | |
let rec loop result tempsum n = | |
printfn "N = %i" n | |
printfn "result = %i" result | |
printfn "tempsum = %i" tempsum | |
if n <=1 then result | |
else | |
let newTempsum = tempsum + n | |
printfn "newTempsum = %i" newTempsum | |
loop (result + (2*(n-1) * newTempsum)) newTempsum (n-1) | |
loop 0 0 n | |
problem_6 100 |> ignore |
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