Sum of all integers from 1 to n

Question

I'm honestly surprised that this hasn't been done already. If you can find an existing thread, by all means mark this as a duplicate or let me know.

Input

Your input is in the form of any positive integer greater than or equal to 1.

Output

You must output the sum of all integers between and including 1 and the number input.

Example

 In: 5
     1+2+3+4+5 = 15
Out: 15

OEIS A000217 — Triangular numbers: a(n) = binomial(n+1,2) = n(n+1)/2 = 0 + 1 + 2 + ... + n.

Leaderboard

Run the code snippet below to view a leaderboard for this question's answers. (Thanks to programmer5000 and steenbergh for suggesting this, and Martin Ender for creating it.)

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Answer 1

Pyth, 2 bytes

sS

Try it online! Implicit input. S is 1-indexed range, and s is the sum.

Answer 2

Husk, 1 byte

Σ

Try it online!

Builtin! Σ in Husk is usually used to get the sum of all elements of a list, but when applied to a number it returns exactly n*(n+1)/2.

Answer 3

Piet, 161 bytes / 16 codels

You can interpret it with this Piet interpreter or upload the image on this website and run it there. Not sure about the byte count, if I could encode it differently to reduce size.

Scaled up version of the source image:

Explanation

The highlighted text shows the current stack (growing from left to right), assuming the user input is 5:

Input a number and push it onto stack

5

Duplicate this number on the stack

5 5

Push 1 (the size of the dark red area) onto stack

5 5 1

Add the top two numbers

5 6

Multiply the top two numbers

30

The black area makes sure, that the cursor moves down right to the light green codel. That transition pushes 2 (the size of dark green) onto stack

30 2

Divide the second number on the stack by the first one

15

Pop and output the top number (interpreted as number)

[empty]

By inserting a white area, the transition is a nop, the black traps our cursor. This ends execution of the program.

Original file (far too small for here):

Answer 4

Brain-Flak, 16 bytes

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

Try it online!

This is one of the few things that brain-flak is really good at.

Since this is one of the simplest things you can do in brain-flak and it has a lot of visibility, here's a detailed explanation:

# Push the sum of all of this code. In brain-flak, every snippet also returns a
# value, and all values inside the same brackets are summed
(
    # Loop and accumulate. Initially, this snippet return 0, but each time the
    # loop runs, the value of the code inside the loop is added to the result.
    {
        # Push (and also return)...
        (
            # The value on top of the stack
            {}

            # Plus the negative of...
            [
                # 1
                ()
            ]

        # The previous code pushes n-1 on to the stack and returns the value n-1
        )

        # 1
        # This code has no side effect, it just returns the value 1 each loop.
        # This effectively adds 1 to the accumulator
        ()

    # The loop will end once the value on top of the stack is 0
    }

    # Pop the zero off, which will also add 0 to the current value
    {}

# After the sum is pushed, the entire stack (which only contains the sum)
# will be implicitly printed.
)

Answer 5

Oasis, 3 bytes

n+0

Try it online!

How it works

n+0
  0    a(0)=0
n+     a(n)=n+a(n-1)

Answer 6

JavaScript (ES6), 10 bytes

n=>n*++n/2

Example

let f =

n=>n*++n/2

console.log(f(5))

Answer 7

Mathematica, 9 bytes

#(#+1)/2&

Mathematica, 10 bytes

(#^2+#)/2&

Mathematica, 11 bytes

Tr@Range@#&

Mathematica, 12 bytes

i~Sum~{i,#}&

Mathematica, 14 bytes

(by @user71546)

1/2/Beta[#,2]&

Mathematica, 15 bytes

Tr[#&~Array~#]&

Mathematica, 16 bytes

Binomial[#+1,2]&

Mathematica, 17 bytes

(by @Not a tree)

⌊(2#+1)^2/8⌋&

Mathematica, 18 bytes

PolygonalNumber@#&

Mathematica, 19 bytes

#+#2&~Fold~Range@#&

Mathematica, 20 bytes

(by @Not a tree)

f@0=0;f@i_:=i+f[i-1]

Answer 8

Python 2, 24 16 bytes

-8 bytes thanks to FryAmTheEggman.

lambda n:n*-~n/2

Try it online!

Answer 9

x86_64 machine language (Linux), 9 8 bytes

0:   8d 47 01                lea    0x1(%rdi),%eax
3:   f7 ef                   imul   %edi
5:   d1 e8                   shr    %eax
7:   c3                      retq 

To Try it online! compile and run the following C program.

#include<stdio.h>
const char f[]="\x8d\x47\x01\xf7\xef\xd1\xe8\xc3";
int main(){
  for( int i = 1; i<=10; i++ ) {
    printf( "%d %d\n", i, ((int(*)())f)(i) );
  }
}

Thanks to @CodyGray and @Peter for -1.

Answer 10

C# (.NET Core), 10 bytes

n=>n++*n/2

Try it online!

Answer 11

Java (OpenJDK 8), 10 bytes

n->n++*n/2

Try it online!

Answer 12

Jelly, 2 bytes

RS

Try it online!

Explanation

RS

    implicit input
 S  sum of the...
R   inclusive range [1..input]
    implicit output

Gauss sum, 3 bytes

‘×H

Explanation

‘×H

     implicit input
  H  half of the quantity of...
‘    input + 1...
 ×   times input
     implicit output

Answer 13

Octave, 22 19 bytes

Because arithmetic operations are boring...

@(n)nnz(triu(e(n)))

Try it online!

Explanation

Given n, this creates an n×n matrix with all entries equal to the number e; makes entries below the diagonal zero; and outputs the number of nonzero values.

Answer 14

Haskell, 13 bytes

This is the shortest (I thinkthought):

f n=sum[1..n]

Try it online!

Direct, 17 13 bytes

f n=n*(n+1)/2

Thanks @WheatWizard for -4 bytes!

Try it online!

Pointfree direct, 15 bytes

(*)=<<(/2).(+1)

Thanks @nimi for the idea!

Try it online!

Pointfree via sum, 16 bytes

sum.enumFromTo 1

Try it online!

Recursively, 22 18 bytes

f 0=0;f n=n+f(n-1)

Thanks @maple_shaft for the idea & @Laikoni for golfing it!

Try it online!

Standard fold, 19 bytes

f n=foldr(+)0[1..n]

Try it online!

Answer 15

Java (OpenJDK 8), 10 bytes

a->a++*a/2

Try it online!

Took a moment to golf down from n->n*(n+1)/2 because I'm slow.

But this isn't a real Java answer. It's definitely not verbose enough.

import java.util.stream.*;
a->IntStream.range(1,a+1).sum()

Not bad, but we can do better.

import java.util.stream.*;
(Integer a)->Stream.iterate(1,(Integer b)->Math.incrementExact(b)).limit(a).reduce(0,Integer::sum)

I love Java.

Answer 16

APL, 3 bytes

+/⍳

Try it online!

+/ - sum (reduce +), ⍳ - range.

Answer 17

Taxi, 687 bytes

Go to Post Office:w 1 l 1 r 1 l.Pickup a passenger going to The Babelfishery.Go to The Babelfishery:s 1 l 1 r.Pickup a passenger going to Cyclone.Go to Cyclone:n 1 l 1 l 2 r.[a]Pickup a passenger going to Addition Alley.Pickup a passenger going to The Underground.Go to Zoom Zoom:n.Go to Addition Alley:w 1 l 1 r.Pickup a passenger going to Addition Alley.Go to The Underground:n 1 r 1 r.Switch to plan "z" if no one is waiting.Pickup a passenger going to Cyclone.Go to Cyclone:n 3 l 2 l.Switch to plan "a".[z]Go to Addition Alley:n 3 l 1 l.Pickup a passenger going to The Babelfishery.Go to The Babelfishery:n 1 r 1 r.Pickup a passenger going to Post Office.Go to Post Office:n 1 l 1 r.

Try it online!

Un-golfed with comments:

[ n = STDIN ]
Go to Post Office: west 1st left 1st right 1st left.
Pickup a passenger going to The Babelfishery.
Go to The Babelfishery: south 1st left 1st right.
Pickup a passenger going to Cyclone.
Go to Cyclone: north 1st left 1st left 2nd right.

[ for (i=n;i>1;i--) { T+=i } ]
[a]
Pickup a passenger going to Addition Alley.
Pickup a passenger going to The Underground.
Go to Zoom Zoom: north.
Go to Addition Alley: west 1st left 1st right.
Pickup a passenger going to Addition Alley.
Go to The Underground: north 1st right 1st right.
Switch to plan "z" if no one is waiting.
Pickup a passenger going to Cyclone.
Go to Cyclone: north 3rd left 2nd left.
Switch to plan "a".

[ print(T) ]
[z]
Go to Addition Alley: north 3rd left 1st left.
Pickup a passenger going to The Babelfishery.
Go to The Babelfishery: north 1st right 1st right.
Pickup a passenger going to Post Office.
Go to Post Office: north 1st left 1st right.

_{It's 22.6% less bytes to loop than it is to use x*(x+1)/2}

Answer 18

Julia, 10 bytes

n->n*-~n/2

Try it online!

11 bytes (works also on Julia 0.4)

n->sum(1:n)

Try it online!

Answer 19

Starry, 27 22 bytes

5 bytes saved thanks to @miles!

, + +  **       +   *.

Try it online!

Explanation

,             Read number (n) from STDIN and push it to the stack
 +            Duplicate top of the stack
 +            Duplicate top of the stack
  *           Pop two numbers and push their product (n*n)
*             Pop two numbers and push their sum (n+n*n)
       +      Push 2
   *          Pop two numbers and push their division ((n+n*n)/2)
.             Pop a number and print it to STDOUT

Answer 20

05AB1E, 2 bytes

LO

Try it online!

How it works

     #input enters stack implicitly
L    #pop a, push list [1 .. a]
 O   #sum of the list
     #implicit output 

Gauss sum, 4 bytes

>¹*;

Try it online!

How it works

>       #input + 1
 ¹*     #get original input & multiply
   ;    #divide by two 

Answer 21

Check, 5 bytes

:)*$p

Check isn't even a golfing language, yet it beats CJam!

Try it online!

Explanation:

The input number is placed on the stack. : duplicates it to give n, n. It is then incremented with ), giving n, n+1. * multiplies the two together, and then $ divides the result by 2. p prints the result and the program terminates.

Answer 22

Piet + ascii-piet, 16 bytes (4×4=16 codels)

tabR   Smm Amtqa

Try Piet online!

How it works

The code simply follows the border and then stops at the 3-cell L-shaped region. Basically uses the well-known formula, because setting up a loop here costs too many cells due to the necessary roll command.

Command    Stack
inN        [n]
dup 1 +    [n n+1]
* 2 /      [n*(n+1)/2]
outN       []

This layout saves two black cells over the following linear, more straightforward (pun intended) layout:

Piet + ascii-piet, 18 bytes (2×9=18 codels)

tabrsaqtM     a mm

Try Piet online!

Answer 23

MATL, 2 bytes

:s

Try it online!

Not happy smiley.

Answer 24

dc, 7 bytes

d1+*2/p

OR

d2^+2/p

OR

dd*+2/p

Try it online!

Answer 25

Brainfuck, 24 Bytes.

I/O is handled as bytes.

,[[->+>+<<]>[-<+>]<-]>>.

Explained

,[[->+>+<<]>[-<+>]<-]>>.
,                           # Read a byte from STDIN
 [                  ]       # Main loop, counting down all values from n to 1
  [->+>+<<]                 # Copy the i value to *i+1 and *i+2
           >[-<+>]          # Move *i+1 back to i
                  <-        # Move back to i, lower it by one. Because *i+2 is never reset, each iteration adds the value of i to it.
                     >>.    # Output the value of *i+2

Answer 26

C++ (template metaprogramming), 80 bytes (?)

I'm not very sure if it is acceptable because you need to insert input into source, which seems to be permitted for languages like /// only.

template<int N>struct s{enum{v=N+s<N-1>::v};};template<>struct s<1>{enum{v=1};};

Example:

#include <iostream>
int main()
{
    std::cout<<s<10>::v;
    return 0;
}

Answer 27

,,,, 6 bytes

:1+×2÷

Explanation

:1+×2÷

:       ### duplicate
 1+     ### add 1
   ×    ### multiply
    2÷  ### divide by 2

If I implement range any time soon...

rΣ

Answer 28

Bash, 13 bytes

seq -s+ $1|bc

Try it online!

seq generates a sequence. seq 5 generates a sequence of numbers from 1 to 5 with a default increment of 1.

seq with the -s flag uses a string parameter to separate the numbers (the default separator is \n).

So seq -s+ $1 generates numbers from 1 to $1, the first argument, using + as the separator. With an argument of 5, this generates 1+2+3+4+5.

Now this is piped into bc using |bc to calculate the result of this mathematical expression and that value it outputted.

Answer 29

Retina 0.8.2, 13 bytes

.+
$*
1
$`1
1

Try it online! Explanation: The first and last stages are just unary ⇔ decimal conversion. The middle stage replaces each 1 with the number of 1s to its left plus another 1 for the 1 itself, thus counting from 1 to n, summing the values implicitly.

Answer 30

Trilangle, 9 bytes

^{How appropriate that the triangular language is computing triangular numbers.}

?2')2!*:@

Try it on the online interpreter!

Unwraps to this triangular grid:

   ?
  2 '
 ) 2 !
* : @ .

Instructions are executed in this order:

• ?: Read an integer from STDIN
• 2: Duplicate the top of the stack
• ): Increment the top of the stack
• *: Multiply the two values off the top of the stack
• '2: Push the number 2 to the stack
• :: Divide the two values off the top of the stack
• !: Print the top of the stack as a decimal number
• @: Terminate program.

Effectively identical to this C program:

#include <stdio.h>

int main() {
  // ?
  int i; while(!scanf("%i", &i)) getchar();

  // 2)*
  i *= i + 1;

  // '2:
  i /= 2;

  // !
  printf("%d\n", i);

  // @
  return 0;
}