Jumlah kuadrat dari sepuluh bilangan alami pertama adalah, $1^2 + 2^2 + \dots + 10^2 = 385$

Kuadrat dari jumlah sepuluh bilangan alami pertama adalah,

$(1 + 2 + ... + 10)^2 = 55^2 = 3025$

Oleh karena itu perbedaan antara jumlah kuadrat dari sepuluh bilangan alami pertama dan kuadrat dari jumlah tersebut

$3025 − 385 = 2640$

Untuk input n yang diberikan, temukan perbedaan antara jumlah kuadrat dari n bilangan alami pertama dan kuadrat dari jumlah.

Uji kasus

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

Tantangan ini pertama kali diumumkan di Project Euler # 6 .

Kriteria Menang

• Tidak ada aturan tentang apa yang seharusnya menjadi perilaku dengan input negatif atau nol.

• Jawaban terpendek menang.

Jelly ,  5  4 byte

Ḋ²ḋṖ

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Bagaimana?

Menerapkan $\sum_{i=2}^n{(i^2(i-1))}$ ...

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  27 byte

-1 terima kasih kepada xnor

lambda n:(n**3-n)*(n/4+1/6)

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

Python 2,  29  28 byte:lambda n:(n**3-n)*(3*n+2)/12

APL (Dyalog Unicode) , 10 byte

1⊥⍳×⍳×1-⍨⍳

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Bagaimana itu bekerja

1⊥⍳×⍳×1-⍨⍳
  ⍳×⍳×1-⍨⍳  Compute (x^3 - x^2) for 1..n
1⊥          Sum

Menggunakan fakta bahwa "kuadrat jumlah" sama dengan "jumlah kubus".

TI-Basic (seri TI-83), 12 11 byte

sum(Ans² nCr 2/{2,3Ans

Implements $\binom{n^2}{2}(\frac12 + \frac1{3n})$. Membawa inputAns: misalnya, jalankan10:prgmXuntuk menghitung hasil untuk input10.

Brain-Flak , 74 72 68 64 byte

((([{}])){({}())}{})([{({}())({})}{}]{(({}())){({})({}())}{}}{})

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Cara yang cukup sederhana untuk melakukannya dengan beberapa perubahan yang rumit. Semoga seseorang akan menemukan beberapa trik untuk membuat ini lebih pendek.

Arang , 12 10 byte

ＩΣＥＮ×ιＸ⊕ι²

$( \sum_1^n x )^2 = \sum_1^n x^3$$( \sum_1^n x )^2 - \sum_1^n x^2 = \sum_1^n (x^3 - x^2) = \sum_1^n (x - 1)x^2 = \sum_0^{n-1} x(x + 1)^2$

   Ｎ        Input number
  Ｅ         Map over implicit range i.e. 0 .. n - 1
        ι   Current value
       ⊕    Incremented
         ²  Literal 2
      Ｘ     Power
     ι      Current value
    ×       Multiply
 Σ          Sum
Ｉ           Cast to string
            Implicitly print

Perl 6 , 22 byte

{sum (1..$_)>>²Z*^$_}

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Menggunakan konstruksi $\sum_{i=1}^n {(i^2(i-1))}$

Japt -x, 9 8 5 4 byte

õ²í*

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Penjelasan

õ        :Range [1,input]
 ²       :Square each
  í      :Interleave with 0-based indices
   *     :Reduce each pair by multiplication
         :Implicit output of the sum of the resulting array

JavaScript, 20 bytes

f=n=>n&&n*n*--n+f(n)

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APL(Dyalog), 17 bytes

{+/(¯1↓⍵)×1↓×⍨⍵}⍳

(Much longer) Port of Jonathan Allan's Jelly answer.

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APL (Dyalog), 16 bytes

((×⍨+/)-(+/×⍨))⍳

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 (×⍨+/)            The square (× self) of the sum (+ fold)
       -           minus
        (+/×⍨)     the sum of the square
(             )⍳   of [1, 2, … input].

Mathematica, 21 17 bytes

^{-4 bytes thanks to alephalpha.}

(3#+2)(#^3-#)/12&

Pure function. Takes an integer as input and returns an integer as output. Just implements the polynomial, since Sums, Ranges, Trs, etc. take up a lot of bytes.

Java (JDK), 23 bytes

n->(3*n+2)*(n*n*n-n)/12

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dc, 16 bytes

?dd3^r-r3*2+*C/p

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

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05AB1E, 8 bytes

ÝDOnsnO-

Explanation:

ÝDOnsnO-     //Full program
Ý            //Push [0..a] where a is implicit input
 D           //Duplicate top of stack
  On         //Push sum, then square it
    s        //Swap top two elements of stack
     nO      //Square each element, then push sum
       -     //Difference (implicitly printed)

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C, C++, 46 40 37 bytes ( #define ), 50 47 46 bytes ( function )

-1 byte thanks to Zacharý

-11 bytes thanks to ceilingcat

Macro version :

#define F(n)n*n*~n*~n/4+n*~n*(n-~n)/6

Function version :

int f(int n){return~n*n*n*~n/4+n*~n*(n-~n)/6;}

Thoses lines are based on thoses 2 formulas :

Sum of numbers between 1 and n = n*(n+1)/2
Sum of squares between 1 and n = n*(n+1)*(2n+1)/6

So the formula to get the answer is simply (n*(n+1)/2) * (n*(n+1)/2) - n*(n+1)*(2n+1)/6

And now to "optimize" the byte count, we break parenthesis and move stuff around, while testing it always gives the same result

(n*(n+1)/2) * (n*(n+1)/2) - n*(n+1)*(2n+1)/6 => n*(n+1)/2*n*(n+1)/2 - n*(n+1)*(2n+1)/6 => n*(n+1)*n*(n+1)/4 - n*(n+1)*(2n+1)/6

Notice the pattern p = n*n+1 = n*n+n, so in the function, we declare another variable int p = n*n+n and it gives :

p*p/4 - p*(2n+1)/6

For p*(p/4-(2*n+1)/6) and so n*(n+1)*(n*(n+1)/4 - (2n+1)/6), it works half the time only, and I suspect integer division to be the cause ( f(3) giving 24 instead of 22, f(24) giving 85200 instead of 85100, so we can't factorize the macro's formula that way, even if mathematically it is the same.

Both the macro and function version are here because of macro substitution :

F(3) gives 3*3*(3+1)*(3+1)/4-3*(3+1)*(2*3+1)/6 = 22
F(5-2) gives 5-2*5-2*(5-2+1)*(5-2+1)/4-5-2*(5-2+1)*(2*5-2+1)/6 = -30

and mess up with the operator precedence. the function version does not have this problem

JavaScript (ES6), 22 bytes

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

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Pyth, 7 bytes

sm**hdh

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Uses the formula in Neil's answer.

sm**hdhddQ   Implicit: Q=eval(input())
             Trailing ddQ inferred
 m       Q   Map [0-Q) as d, using:
    hd         Increment d
   *  hd       Multiply the above with another copy
  *     d      Multiply the above by d
s            Sum, implicit print 

SNOBOL4 (CSNOBOL4), 70 69 bytes

 N =INPUT
I X =X + N ^ 3 - N ^ 2
 N =GT(N) N - 1 :S(I)
 OUTPUT =X
END

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Pari/GP, 21 bytes

n->(3*n+2)*(n^3-n)/12

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05AB1E, 6 bytes

LnDƶαO

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Explanation

L         # push range [1 ... input]
 n        # square each
  D       # duplicate
   ƶ      # lift, multiply each by its 1-based index
    α     # element-wise absolute difference
     O    # sum

Some other versions at the same byte count:

L<ān*O
Ln.āPO
L¦nā*O

R, 28 bytes

x=1:scan();sum(x)^2-sum(x^2)

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MathGolf, 6 bytes

{î²ï*+

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Calculates $\sum_{k=1}^n (k^2(k-1))$

Explanation:

{       Loop (implicit) input times
 î²     1-index of loop squared
    *   Multiplied by
   ï    The 0-index of the loop
     +  And add to the running total

Clojure, 58 bytes

(fn[s](-(Math/pow(reduce + s)2)(reduce +(map #(* % %)s))))

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Edit: I misunderstood the question

Clojure, 55, 35 bytes

#(* %(+ 1 %)(- % 1)(+(* 3 %)2)1/12)

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cQuents, 17 15 bytes

b$)^2-c$
;$
;$$

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Explanation

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

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

APL(NARS), 13 chars, 26 bytes

{+/⍵×⍵×⍵-1}∘⍳

use the formula Sum'w=1..n'(ww(w-1)) possible i wrote the same some other wrote + or - as "1⊥⍳×⍳×⍳-1"; test:

  g←{+/⍵×⍵×⍵-1}∘⍳
  g 0
0
  g 1
0
  g 2
4
  g 3
22
  g 10
2640

Stax , 4 byte

╡⌠(♠

Jalankan dan debug itu

Untuk semua kbilangan bulat positif hingga input, tambahkan k^2 * (k-1).

QBASIC, 45 44 byte

Menghasilkan matematika murni menghemat 1 byte!

INPUT n
?n^2*(n+1)*(n+1)/4-n*(n+1)*(2*n+1)/6

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Sebelumnya, jawaban berbasis loop

INPUT n
FOR q=1TO n
a=a+q^2
b=b+q
NEXT
?b^2-a

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Perhatikan bahwa REPL sedikit lebih diperluas karena penerjemah gagal sebaliknya.

JAEL , 13 10 byte

#&àĝ&oȦ

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Penjelasan (dihasilkan secara otomatis):

./jael --explain '#&àĝ&oȦ'
ORIGINAL CODE:  #&àĝ&oȦ

EXPANDING EXPLANATION:
à => `a
ĝ => ^g
Ȧ => .a!

EXPANDED CODE:  #&`a^g&o.a!

COMPLETED CODE: #&`a^g&o.a!,

#          ,            repeat (p1) times:
 &                              push number of iterations of this loop
  `                             push 1
   a                            push p1 + p2
    ^                           push 2
     g                          push p2 ^ p1
      &                         push number of iterations of this loop
       o                        push p1 * p2
        .                       push the value under the tape head
         a                      push p1 + p2
          !                     write p1 to the tapehead
            ␄           print machine state

05AB1E , 6 byte

LDOšnÆ

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Penjelasan:

           # implicit input (example: 3)
L          # range ([1, 2, 3])
 DOš       # prepend the sum ([6, 1, 2, 3])
    n      # square each ([36, 1, 4, 9])
     Æ     # reduce by subtraction (22)
           # implicit output

Ætidak sering berguna, tetapi ini saatnya untuk bersinar. Ini mengalahkan naif LOnILnO-dengan dua byte penuh.