Problem

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.

Answer

25164150

---

Python

#!/usr/bin/env python
print(sum(range(1, 101))**2 - sum(x**2 for x in range(1, 101)))

---

JavaScript

let sum = 0, sumSquares = 0
for (let i = 1; i <= 100; i++) {
  sum += i
  sumSquares += i * i
}
console.log(sum * sum - sumSquares)

---

Go

package main

import "fmt"

func main() {
	sumSquares, squareSum := 0, 0
	for i := 1; i <= 100; i++ {
		sumSquares += i * i
		squareSum += i
	}
	squareSum *= squareSum
	fmt.Println(squareSum - sumSquares)
}

---

Ruby

#!/usr/bin/env ruby
puts ((1..100).inject(0) {|s,v| s += v})**2 - ((1..100).collect {|x| x**2}.inject(0) { |s,v| s += v})

---

Haskell

main ::  IO ()
main = print $ (s*s) - sqS where
    s = sum [1..100]
    sqS = sum [i * i | i <- [1..100]]

---

Clojure

#!/usr/bin/env clojure
(defn square [x]
  (* x x))

(defn sum-squares [limit]
  (apply + (map square (range 1 (+ limit 1)))))

(defn square-sum [limit]
  (square (apply + (range 1 (+ limit 1)))))

(println (- (square-sum 100) (sum-squares 100)))

---

Rust

fn main() {
    let sum_squares = (1..101).map(|x| x*x).sum::<u64>();
    let square_sum = (1..101).sum::<u64>().pow(2);
    println!("{}", square_sum - sum_squares);
}

---

Mathemtica

(* 
 * Computers are fast, so we can implement this solution directly without any clever math.
 * However for the mathematically inclined, there are closed-form formulas:
 *   sum  = n(n + 1) / 2.
 *   sum2 = n(n + 1)(2n + 1) / 6.
 * Hence sum^2 - sum2 = (n^4 / 4) + (n^3 / 6) - (n^2 / 4) - (n / 6).
 *)
n = 100;
Sum[k, {k, n}]^2 - Sum[k^2, {k, n}]

---

Java

public final class p006 implements EulerSolution {
	
	public static void main(String[] args) {
		System.out.println(new p006().run());
	}
	
	
	/* 
	 * Computers are fast, so we can implement this solution directly without any clever math.
	 * Note that sum^2 is bounded above by (100*100)^2 and sum2 is
	 * bounded above by 100*(100^2), both of which fit in a Java int type.
	 * 
	 * However for the mathematically inclined, there are closed-form formulas:
	 *   sum  = N(N + 1) / 2.
	 *   sum2 = N(N + 1)(2N + 1) / 6.
	 * Hence sum^2 - sum2 = (N^4 / 4) + (N^3 / 6) - (N^2 / 4) - (N / 6).
	 */
	private static final int N = 100;
	
	public String run() {
		int sum = 0;
		int sum2 = 0;
		for (int i = 1; i <= N; i++) {
			sum += i;
			sum2 += i * i;
		}
		return Integer.toString(sum * sum - sum2);
	}
	
}