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

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

$3025 − 385 = 2640$

For a given input n, find the difference between the sum of the squares of the first n natural numbers and the square of the sum.

Test cases

1 => 0
2 => 4
3 => 22
10 => 2640
24 => 85100
100 => 25164150

This challenge was first announced at Project Euler #6.

Winning Criteria

• There are no rules about what should be the behavior with negative or zero input.




• The shortest answer wins.

Jelly,  5  4 bytes

Ḋ²ḋṖ

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How?

Implements $sum_i=2^n(i^2(i-1))$...

Ḋ²ḋṖ - Link: non-negative integer, n
Ḋ - dequeue (implicit range) [2,3,4,5,...,n]
 ² - square (vectorises) [4,9,16,25,...,n*n]
 Ṗ - pop (implicit range) [1,2,3,4,...,n-1]
 ḋ - dot product 4*1+9*2+16*3+25*4+...+n*n*(n-1)

Python 3, 28 bytes

lambda n:n*~-n*-~n*(n/4+1/6)

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Implements $n(n-1)(n+1)(3n+2)/12$

Python 2, 29 bytes: lambda n:n*~-n*-~n*(3*n+2)/12

cQuents, 17 bytes

1$)^2-2$
;$
;$$

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Explanation

 1$)^2-2$ First line
: Implicit (output nth term in sequence)
 1$) Each term in the sequence equals the second line at the current index
 ^2 squared
 -2$ minus the third line at the current index

;$ Second line - sum of integers up to n
;$$ Third line - sum of squares up to n

JavaScript (ES6), 22 bytes

n=>n*~-n*-~n*(n/4+1/6)

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SNOBOL4 (CSNOBOL4), 70 bytes

 N =INPUT
 X =(N * N + N) / 2
 OUTPUT =X ^ 2 - X * (2 * N + 1) / 3
END

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JAEL, 7 chars, 13 bytes

nàṕĝ¯ĝE

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Explanation (generated automatically):

./jael -u explain 'nàṕĝ¯ĝE'
ORIGINAL CODE:
nàṕĝ¯ĝE

EXPANDING EXPLANATION: 
à => `a
ṕ => ´p
ĝ => ^g
ĝ => ^g
E => e!

EXPANDED CODE:
n`a´p^g¯^ge!

COMPLETED CODE:
n`a´p^g¯^ge!

n push [0...p1]
 ` push 1
 a push p1 + p2
 ´ duplicate p1
 p push the sum of all values of set(p1)
 ^ push 2
 g push p2 ^ p1
 ¯ invert p1 and p2 on the stack
 ^ push 2
 g push p2 ^ p1
 e push p2 - p1
 ! write p1 to the tapehead
 ␄ print machine state

Brain-Flak, 74 72 68 bytes

([((([]))(()))(())()]()((()))()(()))

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Pretty simple way of doing it with a couple of tricky shifts. Hopefully someone will find some more tricks to make this even shorter.

I Could not enjoy this app.
Is there anyone here to help me?