Difference of the square of the sum

Question

Find the difference between the square of the sums and sum of the squares.

This is the mathematical representation:

\$\left(\sum n\right)^2-\sum n^2\$

Your program/method should take two inputs, these are your lower and upper limits of the range, and are inclusive. Limits will be whole integers above 0.

Your program/method should return the answer.

You may use whichever base you would like to, but please state in your answer which base you have used.

Test case (Base 10)

5,9      970
91,123   12087152
1,10     2640

This is usual code-golf, so the shorter the answer the better.

Answer 1

Julia, 25 bytes

f(a,b,x=a:b)=sum(x)^2-x'x

This is a function that accepts two integers and returns a 1x1 integer array.

The approach is simple: Construct a UnitRange from the endpoints a and b and call it x, then sum x, square it, and subtract its norm, which is computed as transpose(x) * x.

Try it online! (includes all test cases)

Answer 2

TI-BASIC, 19 bytes

Prompt N,M
randIntNoRep(N,M
sum(Ans)2-sum(Ans2

randIntNoRep gets the range (shuffled). The rest is pretty self explanatory.

Answer 3

Fith, 52 bytes

{ 1 + range dup sum 2 pow swap { 2 pow } map sum - }

This is an anonymous function that takes the two numbers on the stack and leaves a single number.

Explanation:

{
    1 + range dup      2 ranges from a to b inclusive
    sum 2 pow          Sum one and square it
    swap               Bring a fresh range to the top
    { 2 pow } map sum  Square every element and sum the list
    -                  Subtract
}

Answer 4

GeoGebra, 91 bytes

a(x)=(x²+x)/2
b(x)=x³/3+x²/2+x/6
c(x,y)=(a(y)-a(x))²
d(x,y)=b(y)-b(x)
c(x-1,y)-d(x-1,y)

Defines a function (probably e(x,y)) that computes the desired difference.
a(x) calculates the sum of natural numbers between 0 and x.
b(x) calculates the sum of the squares of the natural numbers between 0 and x.
c(x,y) first computes the sum of the natural numbers between x and y, then squares that sum.
d(x,y) calculates the sum of squares between b(x) and b(y).
The last line defines a multi-variable function that finishes the calculation. The function is automatically assigned a name, saving a few bytes.

Answer 5

R - 33 bytes

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

pass your lower limit to the first scan then your upper limit to the second scan

Answer 6

Whispers v2, 90 bytes

> Input
> Input
> 2
>> 1…2
>> L*3
>> Each 5 4
>> ∑6
>> ∑4
>> 8*3
>> 9-7
>> Output 10

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Answer 7

Julia 0.6, 21 bytes

r->sum(r)^2-sum(r.^2)

Returns an anonymous function that takes a Range, eg f(1:10), which then returns the result.

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Answer 8

C (gcc), 67 bytes

f(n,m){double a,b;for(;n<=m;a+=n,b+=pow(n++,2));return pow(a,2)-b;}

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Answer 9

C (gcc), 48 bytes

a,b;f(n,m){for(a=b=0;m/n;a+=n++)b+=n*n;n=a*a-b;}

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Answer 10

Red, 70 bytes

func[a b][s: t: 0 until[s: s + a t: a * a + t  b < a: a + 1]s * s - t]

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Answer 11

Excel, 60 bytes

=(B1^2+B1-A1^2+A1)^2/4-(2*B1^3+3*B1^2+B1-2*A1^3+3*A1^2-A1)/6

Algebraic reshuffling of more verbose implementation: For inputs a and b:

(B1*(B1+1)-A1*(A1-1))/2          // (Sum of b) - (Sum of a-1)
(  )^2                           //  squared
B1*(B1+1)*(2*B1+1)/6             //  Sum of (b squared)
A1*(A1-1)*(2*A1-1)/6             //  Sum of (a-1 squared)

Answer 12

C# (.NET Core), 62 bytes

(a,b)=>{int x=0,y=0;for(;a<=b;a++){x+=a;y+=a*a;}return x*x-y;}

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Ungolfed:

(a, b) => {                 // takes in two integer inputs, delimited by a comma
    int x = 0, y = 0;       // initializes x and y
    for(; a <= b; a++)      // increment a until a equals b
    {
        x += a;             // add current a to x
        y += a * a;         // add square of current a to y
    }
    return (x * x) - y;     // return the difference of the square of the sums (x*x) and the sum of the squares (y)
}

Answer 13

PHP, 58 bytes

for([,$p,$q]=$argv;$p<=$q;$s+=$p++)$t+=$p*$p;echo$s*$s-$t;

Run with -nr or try it online.

Answer 14

Python 3, 66 65 bytes

lambda a,b:sum(x*y*2for x in range(a,b+1)for y in range(x+1,b+1))

Try it online!

with @Dennis formulae and w/o recursion, yet another Python solution (as I cannot comment @leaky-nun solution (not enough reputation)

Answer 15

Sidef, 46 44 bytes

Translation of: my Factor solution

{|a,b|R=a..b;R.map{.sqr}.sum-R.sum.sqr->abs}

Try it online!

use it like x.run(4, 20) -> 38760

Previous

->(a,b,R=a..b){R.map{.sqr}.sum-R.sum.sqr->abs}

longer, explained

->                       # the next list is a function parameter list
(a, b, R = a .. b )      # R is a RangeNum over the a and b 
{                        # a Block
  return                 # return the whole expression
  R.map{ |n|             # map this Block over the range
    return n.sqr         # squared value
  }.sum                  # summed the range
  -                      # subtract
  R.sum.sqr              # sum of the range, squared
  ->abs                  # call Number.abs on everything on the left-hand side of ->
}                        # ok, we're done 

Answer 16

Burlesque - 16 bytes

r@J++2?^j)S[++?-

r@                  range
  J                 dup
   ++               sum
     2?^            square it
        j           swap
         )S[        map (square)
            ++      sum
              ?-    subtract

Alternate versions:

r@JJ?*++j++S[?-ab
r@J++2?^jqS[ms?-
r@J++J?*jqS[ms?-
r@J++J?*jJ?*++?-
r@J)S[++j++S[j?-

^- and this is the art of golfing :(.

Answer 17

Rust, 67 bytes

|b,e|{let mut t=0;let mut s=0;for i in b..=e{t+=i;s+=i*i;}(t*t)-s};

A closure that takes two integers and returns an integer.

Answer 18

Tcl, 76 80 bytes

proc S l\ u {while \$l<=$u {incr Q $l
append R -$l*$l
incr l}
expr $Q**2+$R}

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Answer 19

Pony, 73 bytes

fun f(m:U64,n:U64,s:U64=0):U64=>if m>n then 0 else(2*m*s)+f(m+1,n,m+s)end

sort of ungolfed

  fun apply(m: U64, n: U64, s: U64 = 0): U64 =>
    if m > n then 0
    else (2 * m * s) + apply(m + 1, n, m + s) end

Port of Neil's JavaScript solution.

Answer 20

K (oK), 24 bytes

{(t*t:+/s)-+/s*s:x_!1+y}

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"Boring" but straightforward.

Answer 21

JavaScript, 49 bytes

Without recursion:

(b,e)=>eval('for(x=y=0;b<=e;y+=b*b++)x+=b;x*x-y')

Try it:

f=(b,e)=>eval('for(x=y=0;b<=e;y+=b*b++)x+=b;x*x-y')

console.log(f(5,9)); // 970
console.log(f(91,123)); // 12087152
console.log(f(1,10)); // 2640

Answer 22

Pyt, 8 bytes

ŘĐĐƩ²↔·-

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Takes two inputs on separate lines as max, min

Ř               implicit inputs; Řange
 ĐĐ             Đuplicate array on stack twice
   Ʃ²           Get square of sum
     ⇹·         Get sum of squares
       -        Subtract; implicit print

Answer 23

Ly, 22 bytes

nnR&s&+:*>l[:*sp<l->]<

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This is a straight forward interpretation of the rules...

nnR                     - get the start/end, generate the sequence of numbers
   &s                   - save the sequence...
     &+                 - sum it
       :*               - square the sum
         >l             - load the sequence on a new stack
           [        ]   - while the stack isn't empty
            :*          - square the number
              sp<l      - move the squared number to the other stack
                  ->    - subtract and go back to the second stack
                     <  - switch back to the accumulator stack
                        - the number on the stack prints automatically on exit

Answer 24

Arturo, 42 35 bytes

$=>[-*<=∑r∑map r:<=&..&'z->z*z]

Try it

Answer 25

J-uby, 33 bytes

:!~|:-%[:sum|g= ~:**&2,:*&g|:sum]

(The space between = and ~ is necessary to keep the Ruby lexer from interpreting them as the =~ operator.)

Explanation

:!~ | :- % [:sum | g = ~:** & 2, :* & g | :sum]

:!~ |                                            # Range, then
      :- % [                   ,              ]  # difference of...
            :sum | g = ~:** & 2                  #   sum, then square (save the square function in g), and
                                 :* & g | :sum   #   map with square, then sum

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Answer 26

Java, 95 bytes

public int d(int a,int b){int i=a,j,s=0;for(;i<=b;i++)for(j=a;j<=b;j++)s+=i==j?0:i*j;return s;}

Ideone it!

Answer 27

Python, 62 bytes

lambda a,b:sum(range(a,b+1))**2-sum(x*x for x in range(a,b+1))

Or, longer:

lambda a,b:sum(sum(x*y for y in range(a,b+1)if x-y)for x in range(a,b+1))

Ideone both!

Answer 28

Actually, 11 bytes

u@x;♂²Σ@Σ²-

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Answer 29

R, 23 bytes

sum(i:j)^2-sum((i:j)^2)

Does pretty much what you expect by taking i and j as inputs.

Answer 30

Actually, 10 bytes

u@x;;*@Σ²-

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Explanation:

u@x;;*@Σ²-
u@x         range(a, b+1)
   ;;       two copies
     *      dot product with itself (equal to sum of squares)
      @Σ²   sum remaining copy, square sum
         -  subtract