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

Neim, 3 2 bytes

First Neim answer

𝐈𝐬

Try it online!

Explanation

𝐈   # Gets inclusive range from 0 to input
𝐬   # Sum the list

Saved a byte due to Okx

Answer 2

C, 56 47 bytes

main(n){scanf("%d",&n);printf("%d",n*(n+1)/2);}

This is my first attempt at any code golf of any kind. Submitted as I saw that there were no other answers for C.

Old code:

int main(){int n;scanf("%d",&n);printf("%d",n*(n+1)/2);}

Thanks to programmer5000 for the help. :)

Answer 3

MUMPS, 15 bytes

r n w !,n*n+n/2

Accepts user input (r n) and writes a new line along with the sum (w !,n*n+n/2). Order of operations doesn't matter in MUMPS: It goes from left to right except when there are parentheses.

Answer 4

Excel, 12 bytes

=(A1+1)/2*A1

Or, alternatively:

=(A1^2+A1)/2

Instead of counting all n elements, take the average of the n elements, and multiply it by the number of elements.

Answer 5

R, 13 bytes

sum(1:scan())

Saved 9 bytes thanks to Giuseppe and Max Lawnboy.

Answer 6

Forth, 17 bytes

Defines a word (function) that returns n*(n+1)/2.

: f dup 1+ * 2/ ;

Try it online

---

Full program with the same byte count:

key dup 1+ * 2/ .

Try it online - input is a single character, like BF.

Answer 7

C, 22 bytes

With preprocessor

#define F(n) (n+1)*n/2

or 24 bytes with code and math

F(n){return (n+1)*n/2;}

or 27 bytes with recursive code

F(n){return n?n+F(n-1):0;}

Answer 8

ArnoldC, 461 bytes

IT'S SHOWTIME
HEY CHRISTMAS TREE I
YOU SET US UP 0
GET YOUR ASS TO MARS I
DO IT NOW
I WANT TO ASK YOU A BUNCH OF QUESTIONS AND I WANT TO HAVE THEM ANSWERED IMMEDIATELY
HEY CHRISTMAS TREE Z
YOU SET US UP I
GET TO THE CHOPPER I
HERE IS MY INVITATION I
GET UP 1
ENOUGH TALK
GET TO THE CHOPPER Z
HERE IS MY INVITATION Z
YOU'RE FIRED I
ENOUGH TALK
GET TO THE CHOPPER Z
HERE IS MY INVITATION Z
HE HAD TO SPLIT 2
ENOUGH TALK
TALK TO THE HAND Z
YOU HAVE BEEN TERMINATED

Explanation

Schwarzy.

Try it online!

Answer 9

J, 6 4 bytes

2!>:

Edit: I forgot that the binomial coefficient formula existed, so that lowers the bytecount. Also this is on the REPL or as a function with the input taken as the right argument. The other solutions need to be on the REPL, which I forgot to mention.

First post in a while, figured I'd submit the language I've been trying to learn recently. Not sure if you can specify one-indexing for ranges in J like with APL.

Explanation

2!>:
  >:  Increment
2!    n Choose 2

Previous solution (6 bytes)

Explanation below

+/i.>:
    >:  Add 1
  i.    Range [0,n+1)
+/      Sum

7 byte solutions

Explanations beneath each

-:(*>:)
  (*>:)  Hook: n * (n+1)
-:       Halve

-:(+*:)
  (+*:)  Hook: n^2 + n
-:       Halve

Answer 10

Röda, 13 bytes

{seq 1,_|sum}

Try it online!

Answer 11

Vim, 4̶6̶ 25 17 16 15 keystrokes

Thanks @CowsQuack for -9 bytes!

YP<C-a>Jr*0C<C-r>=<C-r>"/2⏎

Try it online!

Ungolfed/Explained

YP                           " duplicate line containing N
  <C-a>                      " increment the first line
       J                     " join the two lines
        r*                   " substitute space between N and N+1 with *
          0C                 " delete line (store in " register) and insert
            <C-r>=        ⏎  "   the expression
                  <C-r>"     "   from the " register
                        /2   "   divided by 2

Answer 12

ArnoldC, 301 bytes

Well if Leo can find a way to do it in one byte, this is my way of throwing in the towel.

With the language based on the guy who never surrenders.

(And studying for the precalc final, you know, n(n+1)/2 is a formula I won't forget now, right?)

As of now, there's not really a way to take input in from the console from Try It Online, but this guy supposedly added something here.

Assuming that works, this code should do:

IT'S SHOWTIME
HEY CHRISTMAS TREE n
YOU SET US UP 0
GET YOUR ASS TO MARS n
DO IT NOW
WHO IS YOUR DADDY AND WHAT DOES HE DO
HEY CHRISTMAS TREE a
YOU SET US UP n
GET TO THE CHOPPER a
HERE IS MY INVITATION a
GET UP 1
YOU'RE FIRED n
HE HAD TO SPLIT 2
ENOUGH TALK
TALK TO THE HAND a
YOU HAVE BEEN TERMINATED

If not, this should work, manually assigning variable n (although it's a bit against the challenge)

IT'S SHOWTIME
HEY CHRISTMAS TREE n
YOU SET US UP 3
HEY CHRISTMAS TREE a
YOU SET US UP n
GET TO THE CHOPPER a
HERE IS MY INVITATION a
GET UP 1
YOU'RE FIRED n
HE HAD TO SPLIT 2
ENOUGH TALK
TALK TO THE HAND a
YOU HAVE BEEN TERMINATED

Try it online!

Answer 13

Pyt, 1 byte

△

Try it online!

Explanation:

               implicit input
 △             compute the nth triangle number

Answer 14

brainfuck, 14 bytes

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

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Takes input as character code, outputs as unary null bytes

Answer 15

shortC,  44  29 bytes

Bn){K"%d",&n);R"%d",n*(n+1)/2

Try it online!

Just a shortCed version of this. Any help would be appreciated.

Answer 16

Spice (56 bytes)

;a;b;c;d@
REA a;
ADD a 1 b;
DIV b 2 c;
MUL a c d;
OUT d;

Explanation

The code is an implementation of the n*(n+1)/2 solution seen in other responses. It should be fairly readable but here's the code annotated:

;a;b;c;d@    - Declare 4 vars a, b, c and d (@ marks end of declarations)
REA a;       - Read in the value from the console and store in a
ADD a 1 b;   - Add 1 to a and store in b
DIV b 2 c;   - Divide b by 2 and store in c
MUL a c d;   - Mulitply a and c and store in d
OUT d;       - Output d to console

Answer 17

C# (Visual C# Interactive Compiler), 29 22 10 bytes

n=>n*++n/2

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Answer 18

C (gcc), 19 bytes

#define f(n)n*-~n/2

Try it online!

C (gcc), 15 bytes

f(n){n*=-~n/2;}

-5 bytes thanks to ceilingcat!

Try it online!

Answer 19

Hexagony, 16 bytes

?"&{2'<)}=*/@!:=

Try it online! Try it online differently!

Answer 20

Fourier, 9 bytes

I~_^*_/2o

Try it online!

Answer 21

Knight, 11 bytes

O/*+1=xPx 2

Try it online!

Ungolfed:

OUTPUT / (* (+1 = x PROMPT) x) 2

How the magic works:

This is using the fact that \$\sum_{n=1}^{x} n == \frac{x(x + 1)}{2}\$

1. =xP First, we read a line from standard input and assign to x. Normally, we would do +0P to immediately coerce it to a Number, but I have better plans.
2. +1=xP We add the result of the assignment to \$1\$. This coerces to Number for us, giving a result of Number.
3. *+1=xPx We then multiply that by x. Since +1=xP is a Number, x will also be converted to a Number. Now, we have \$x(x + 1)\$.
4. Last, divide by 2 and output.

Answer 22

MSWLogo, 26 bytes

to f :n
pr :n*(:n+1)/2
end

Answer 23

Vyxal Rs, 0 bytes

Try it Online!

Lol

Answer 24

Desmos, 13 bytes

f(n)=.5nn+.5n

Very surprised that there wasn't a Desmos answer here yet.

Try It On Desmos!

Answer 25

Regex 🐇 (Perl / Raku / PCRE), 2 bytes

x+

Takes its input in unary, as a string of x characters whose length represents the number. Returns its output as the number of ways the regex can match.

Try it online! - Perl v5.28.2 / Attempt This Online! - Perl v5.36+
Try it online! - Raku
Try it online! - PCRE1
Try it online! - PCRE2 v10.33 / Attempt This Online! - PCRE2 v10.40+

Although Perl, Raku, and PCRE2 are the only regex engines currently capable of counting the number of possible matches without their source code being patched, this solution itself obviously only uses the most basic regex functionality, universal to all regex engines.

x+ will match every possible non-empty substring exactly once. So for example with the input 5, it is converted to unary as xxxxx, and then the list of possible matches is the following set of substrings:

xxxxx, xxxx, xxx, xx, x
xxxx, xxx, xx, x
xxx, xx, x
xx, x
x

As such, counting these is equivalent to calculating the sum of all integers from \$1\$ to \$n\$ (a.k.a. the \$n\$th triangular number).

This can be extended in many interesting ways, for example raising numbers to a constant power.

Retina 1.0, 9 bytes

.+
*
w`.+

Input and output are both in decimal.

Try it online!

Explanation:

.+
*

Convert the input from decimal to unary, with _ as the repeating character.

w`.+

w enables full overlap mode, where all substrings are considered (this feature was introduced in Retina 1.0.0). .+ then matches the full set of non-empty substrings. Using .* instead, to match all substrings, would shift the sequence 1 forward: Try it online!

If the input were taken in unary, this solution would be 4 bytes: Try it online!

Retina is not capable of counting the number of ways a regex can match, only the number of unique substrings that match, so this method can't be directly extended any further than the triangular number sequence.

Answer 26

BitCycle -u, 11 10 bytes

-1 byte thanks to Jo King

 ~\~!
?AB^

Try it here!

Explanation

In BitCycle, numbers are represented in unary (with the -u flag allowing I/O to be given in decimal form for convenience). Thus, all we need is this algorithm:

• Input a number
• Loop while the number is nonzero:
  • Output the current number
  • Decrement the current number

Outputting unary numbers consecutively has the effect of adding them, thus giving n + (n-1) + ... + 1.

The number (as a string of 1 bits) comes in at the source ? and goes into the collector A, which then dumps it into the collector B. When the bits exit B, they are directed north into the dupneg ~, where a copy turns right (east) and a negated copy turns left (west). The original copy goes into the sink ! to be output. From the negated copy, one bit is sent north by the splitter \ and discarded. The rest get negated again by a second dupneg; the doubly negated bits go south into the A collector to begin the loop again, while the other copies go north off the playfield and are discarded. Once no bits are left, the program halts and outputs the sink's contents.

Answer 27

flax, 2 bytes

Σκ

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Explanation

Σκ
Σ     ⍝ sum of
 κ    ⍝ numbers from range 1...n

Answer 28

awk, 18 bytes

jot 50 | 

awk '$_*=$++_++/(_+_--)'

1
3
6
10
15
21
28
36
45
55
66
78
91
105

Answer 29

Japt -mx, 1 byte

Ä

Try it

Ä     :Implicit map of each U in the range [0,input)
Ä     :Add 1
      :Implicit output of sum of resulting array

Answer 30

Nibbles, 1 byte (2 nibbles)

+,

No builtin for sum-over-list or flags for summation or range, but Nibbles still seems pretty competitive here...

+       # sum of list:
 ,      # range from 1..
        # (implicit) input