# Sum of all integers from 1 to n

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.

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.)

• Closely related Jul 18, 2017 at 20:36
• @FryAmTheEggman Sorry - had a bit of a brain fart there. I see what you mean. Jul 18, 2017 at 20:45
• @Aaron you got ninja'd by Husk, which was just posted with a 1 byte solution Jul 18, 2017 at 21:35
• I suggest a stack snippet.
– user58826
Jul 19, 2017 at 11:42
• Jul 27, 2017 at 12:20

IEK+O@

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# @yBASIC, 10 bytes

?_*_+_>>!.

Input should be in _ (No input methods exist)

## Explanation

N(N+1) can be rewritten as N*N+N. Dividing by 2 would then require parentheses, but a left shift can be used instead. !. is the same as !0.0 (logical not of 0), which is 1.

# MathGolf, 2 bytes

╒Σ

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Pretty much exactly the sum (Σ) of the range from 1 to input ()

# Gol><>, 7 bytes

I:P*2,h

Courtesy of JoKing

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8 bytes

I::*+2,h

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# Go, 29 bytes

A function literal which uses the closed form, $$\ \frac n2(n+1) \$$.

func(n int)int{return-~n*n/2}

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# Symbolic Python, 14 bytes

_*=-~_*_/(_+_)

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This basically evaluates to $$\ \frac{ n^2 (n-1)}{2n} = \frac{ n (n-1)}{2}\$$, which is the usual form for triangular numbers.

# Shakespeare Programming Language, 154 bytes

S.Ajax,.Puck,.Act I:.Scene I:.[Enter Ajax and Puck]
Ajax:
Listen tothy.
You is the quotient betweenthe product ofyou the sum ofyou a cat a big cat.
Open heart

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Simple, just does n(n+1)/2.

# Charcoal, 5 bytes

ＩΣ…·Ｎ

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### Explanation:

Σ        The sum of
…·Ｎ     the inclusive range from 0 to input
Ｉ          converted to a string

I tried to make it with the formula $$\\frac{x(x+1)}{2}\$$ but it took more characters.

Ï⅀

TIO

## Keg, 5 bytes (SBCS)

Ï∑+).

Explanation:

Ï#    Range from input to 0. The 0 will not affect the summation.
)# We have to complete the braces if we want to output as an integer.
.#Output as an integer

TIO

• That's available on the github interpreter and theoretically, it should work. Oct 13, 2019 at 1:13
• -1 byte using the -hr flag Dec 6, 2019 at 20:43

## Clojure - 19 bytes

#(apply +(range %))

# Labyrinth, 8 bytes

?:)*_2/!

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A linear program which halts by error at the bounced-off / (triggering division by zero).

?:)*  Push input, dup, increment, multiply (giving n*(n+1))
_2/   Divide by 2
!     Pop and print, bounce off the dead end
/     The top is (implicit) 0 so it triggers division by zero error

# MashedPotatoes, 144 bytes

synchronized
(0.0f){casestd::ignoreof{_->usestrictqw/nullptr/;}goto++i;(formatt"WHILE$ARGV<$[SETLOCAL*read-eval*WEND")procFS{uniq-c}{s/()//g}}

MashedPotatoes isn't available on TIO, so here's the Esolangs page for reference.

https://esolangs.org/wiki/MashedPotatoes

The annotated code below might not make sense without reading about the quirky language syntax first, but here goes...

synchronized
(0.0f) {

Sets the value of Label0 to 2 (the line number where the ( appears. Important for register rotation expression later.

case std::ignore of { _ -> use strict qw/nullptr/; }

Read an integer from STDIN into register ^A.

goto ++i;

Shift the registers, meaning ^A -> ^C -> ^E, to move the input to ^C.

(format t "

Loops while the contents of ^C are greater than 0.

WHILE $ARGV <$[ SETLOCAL *read-eval* WEND

Adds the value of ^C to register ^E.

")

End of loop, automatically decrements ^C.

proc FS {uniq -c} { s/()//g }

Prints the contents of ^E.

}

Just closing the synchronized statement to end the program cleanly