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

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 7

JavaScript (ES6), 10 bytes

n=>n*++n/2

Example

let f =

n=>n*++n/2

console.log(f(5))

Answer 8

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 9

Python 2, 24 16 bytes

-8 bytes thanks to FryAmTheEggman.

lambda n:n*-~n/2

Try it online!

Answer 10

C# (.NET Core), 10 bytes

n=>n++*n/2

Try it online!

Answer 11

Octave, 22 19 bytes

Because arithmetic operations are boring...

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

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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 12

Java (OpenJDK 8), 10 bytes

n->n++*n/2

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Answer 13

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 14

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 15

APL, 3 bytes

+/⍳

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+/ - sum (reduce +), ⍳ - range.

Answer 16

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*;

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How it works

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

Answer 17

Starry, 27 22 bytes

5 bytes saved thanks to @miles!

, + +  **       +   *.

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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 18

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 19

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 20

MATL, 2 bytes

:s

Try it online!

Not happy smiley.

Answer 21

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 22

,,,, 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 23

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 24

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 25

dc, 7 bytes

d1+*2/p

OR

d2^+2/p

OR

dd*+2/p

Try it online!

Answer 26

8th, 21 12 bytes

Saved 9 bytes thanks to FryAmTheEggman

dup 1+ * 2 /

Usage and output

ok> : sum dup n:1+ * 2 / ;

ok> 5 sum .
15

Answer 27

><>, 7+3 = 10 bytes

Calculates n(n+1)/2.
3 bytes added for the -v flag

:1+2,*n

Try it online!

Or if input can be taken as a character code:

><>, 9 bytes

i:1+2,*n;

Try it online!

Answer 28

Retina, 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 29

PHP, 19 bytes

<?=$argn*-~$argn/2;
<?=$argn/2*++$argn;
<?=$argn*++$argn/2; # this one fails

using builtins, 29 bytes:

<?=array_sum(range(1,$argn));

loop, 31 bytes:

while($argn)$s+=$argn--;echo$s;

Answer 30

x86-64 Machine Code, 7 bytes

31 C0
01 C8
E2 FC
C3  

The above bytes define a function that accepts a single parameter, n, and returns a value containing the sum of all integers from 1 to n.

It is written to the Microsoft x64 calling convention, which passes the parameter in the ECX register. The return value is left in EAX, like all x86/x86-64 calling conventions.

Ungolfed assembly mnemonics:

       xor  eax, eax    ; zero out EAX
Next:  add  eax, ecx    ; add ECX to EAX
       loop Next        ; decrement ECX by 1, and loop as long as ECX != 0
       ret              ; return, with result in EAX

Try it online!
_{(The C function call there is annotated with an attribute that causes GCC to call it using the Microsoft calling convention that my assembly code uses. If TIO had provided MSVC, this wouldn't be necessary.)}

---

By the unusual standards of code golf, you see that this iterative looping approach is preferable to approaches that use the more sane mathematical formula (n(n+1) / 2), even though it is obviously vastly less efficient in terms of run-time speed.

Using number theory, ceilingcat's implementation can still be beat by one byte. Each of these instructions are essential, but there is a slightly shorter encoding for IMUL that uses EAX implicitly as a destination operand (actually, it uses EDX:EAX, but we can just ignore the upper 32 bits of the result). This is only 2 bytes to encode, down from 3.

LEA takes three bytes as well, but there's really no way around that because we need to increment while preserving the original value. If we did a MOV to make a copy, then INC, we'd be at 4 bytes. (In x86-32, where INC is only 1 byte, we'd be at the same 3 bytes as LEA.)

The final right-shift is necessary to divide the result in half, and is certainly more compact (and more efficient) than a multiplication. However, the code should really be using shr instead of sar, since it's assuming that the input value, n, is an unsigned integer. (That assumption is valid according to the rules, of course, but if you know that the input is unsigned, then you shouldn't be doing a signed arithmetic shift, as the upper bit being set in a large unsigned value will cause the result to be incorrect.)

8D 41 01                lea    eax, [rcx+1]
F7 E9                   imul   ecx
D1 E8                   shr    eax, 1
C3                      ret

Now only 8 bytes (thanks to Peter Cordes). Still, 8 > 7.

Answer 31

Cubix, 12 10 bytes

*,)2I://O@

Initial version

....I:)*2,O@

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Explanation

Expanded onto a cube, the code looks like this:

    * ,
    ) 2
I : / / O @ . .
. . . . . . . .
    . .
    . .

The instruction pointer (IP) starts at the I, moving east. It continues moving east until it comes across the / mirror, which reflects it north. When the IP reaches the top of the code, it wraps around to the last . on the third line, moving south. Then it wraps to the penultimate . on the last line, moving north. Then it reaches the / mirror again, which reflects it east, only for the next / to reflect it north again. This time, the IP wraps to the penultimate . on the third line, and then the last . on the last line.

The instructions are executed in the following order.

I:)*2,O@ # Explanation
I        # Take input as an integer and push it to the stack
 :       # Duplicate the input
  )      # Increment one of the inputs
   *     # Multiply the input by input+1
    2    # Push 2 to the stack
     ,   # Integer devide the multiplication result by 2
      O  # Output the result
       @ # End program

Answer 32

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 33

ArnoldC, 310 bytes

Removes unnecessary variable assignments from Courtois' solution and replaces them with GET TO THE CHOPPER and some arithmetic operations.

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

Who doesn't like some good Arnold Schwarzenegger one liners :)

Try it online!

Answer 34

Ohm, 2 bytes

@Σ

Try it online!

Explanation

@Σ

    implicit input
@   inclusive range [1..input]
 Σ  sum
    implicit output

Answer 35

Ruby, 29 17 18 14 12 bytes

->n{n*-~n/2}

Execution:

->n{n*-~n/2}.call(5)

Gets The Value of the sum of 1 through 5

Try It Out!

People Who Have Saved Me A Few Bytes:

Saved 12 Bytes - DJMcMayhem

Saved 6 Bytes - FryAmTheEggman

Unsaved 3 Bytes (But Added Variable Handling) - Value Ink

Fixed A Misunderstanding, Saving me 8 Bytes - Value Ink

Saved 2 Bytes - G B

Many Thanks!

Answer 36

cQuents, 2 bytes

;$

This is the type of question that cQuents was designed for, and the type of question I implemented the ; mode for. Take that, Oasis!

Try it online!

Explanation

;    Mode: Sum (output sum of sequence up to input)
 $   Each item in the sequence is its (1-based) index

Answer 37

WendyScript, 17 bytes

<<f=>(x)x*(x+1)/2

f(100) // => 5050

Try it online!

Answer 38

Lean Mean Bean Machine, 38 32 bytes

-5 bytes thanks to Roman Gräf
-1 byte from changing LMBM's division peg from £ to ,

 O O
 i 2
 o
  )/
  ,
  /
 /
*
u

Explanation

Each O spawns a marble at program start. The first marble reads input and has it's value set to it, the 2nd has it's value set to 1, and the 3rd has it's value set to 2.

The n-marble is then duplicated, one copy falls all the way to a multiplication operator, where it will be held for a 2nd marble, the other falls into a subtraction operator, which the 1-marble then falls into after it.

This new n-1-marble then falls into a division operator (,), and the 2-marble falls in right after it.

This (n-1)/2-marble then falls into the multiplication operator, and the final n*(n-1)/2 marble falls into a u peg, where its value is printed, and the marble is destroyed.

Answer 39

Bash, 26, 19 bytes

echo $[($1+1)*$1/2]

Try it online!

19 bytes for the code, thanks to rexkogitans.

Answer 40

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 41

Brachylog, 2 bytes

⟦+

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Explanation

⟦      Range: [0, …, Input]
 +     Sum:   0 + … + Input

Answer 42

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 43

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 44

Python 2, 16 bytes

lambda n:-~n*n/2

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Answer 45

Triangular, 10 bytes

$\:_%i/2*<

Ungolfed:

   $
  \ :
 _ % i
/ 2 * <

Try it online!

The code, without directionals, is read as $:i*2_%.

• $ reads an integer x, stack contains {x}.
• : duplicates it, stack contains {x,x}.
• i increments the top of stack, stack contains {x,x+1}.
• * multiplies the top two stack values, stack contains {x*(x+1)}.
• 2 pushes 2 to the stack, stack contains {x*(x+1),2}.
• _ divides the top two stack values, stack contains {x*(x+1)/2}.
• % prints the top of stack, the equation x*(x+1)/2.

^{Idea thanks to caird, who asked me to post.}

Answer 46

TI-BASIC, 6 bytes

Beating Casio-basic by 3 bytes :) 6 byte version thanks to PT_ from cemetech (https://www.cemetech.net/forum/profile.php?mode=viewprofile&u=10064)

mean({N²,N

Two other, 7 byte, versions:

.5N(N+1 

.5(N²+N

Answer 47

TI-Basic, 6 bytes

sum(randIntNoRep(1,Ans

Alternate solutions:

mean({AnsAns,Ans   6 bytes credits to @user1812
.5Ans(Ans+1        7 bytes
.5(AnsAns+Ans      7 bytes
Σ(I,I,1,Ans        9 bytes
sum(seq(I,I,1,Ans  9 bytes

Answer 48

brainfuck, 24 bytes

Input number in base255(ASCII), use bigger cells for larger numbers, outputs also in base255.

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

Try it online!

For bigger cells.

    ,[         Get input in base 255 mod 2^bit
         [ >+  Copy it left(to preserve index) 
           >+  and left left to accumulate the sum
     <<- ]     decrement index to break loop
         >     Move to the first copy, index'in
         [-<+>]Move it back, restoring the index
  <- ]         Decrement index, let function run again until 0
    >>. Print sum 

Answer 49

Symbolic Python, 35 33 18 16 bytes

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

Try it online!

Uses the direct formula for triangle numbers, (n+1)(n/2):

_=                  # Set output to
  -~_               #   (n+1)
     *_             #   *n
       /-~(_==_)    #   /2

Answer 50

Alchemist, 40 bytes

0x+_->a+Out_I
0_+0x+a->x
x+a->_+x
x+0a->

Outputs in unary, try it online or try it with automatic conversion to decimal!

Explanation

We'll 3 types of atoms the $\texttt{_}$-, $\texttt{x}$- and $\texttt{a}$-atoms:

• $\texttt{_}$ initially is the input
• $\texttt{x}$ is to make sure the computation is deterministic
• $\texttt{a}$

# When there is no x-atom but still _ left, transform it to a and output I
0x + _ -> a + Out_I
# If there are no _- and x-atoms but still at least one a, we remove one a and add an x
0_ + 0x + a -> x
# If there is an x-atom, exchange all a-atoms for _-atoms
x + a -> _ + x
# Once we're done with that, remove the x-atom (this makes the first rule applicable again, creating a loop)
x + 0a ->

So essentially we transform each $\texttt{_}$ to $\texttt{a}$ while outputting an $\texttt{I}$, remove one $\texttt{a}$ and exchange them back to $\texttt{_}$, until there are no $\texttt{a}$-atoms and no rule is applicable anymore - terminating the progress of the universe.

Answer 51

PowerShell, 22 18 bytes

param($n)$n*++$n/2

Try it online!

Saved 4 bytes thanks to FryAmTheEggman. Uses Gauss' formula. Ho-hum.

Answer 52

R, 13 bytes

sum(1:scan())

Saved 9 bytes thanks to Giuseppe and Max Lawnboy.

Answer 53

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 54

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 55

Neim, 3 2 bytes

First Neim answer

Try it online!

Explanation

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

Saved a byte due to Okx

Answer 56

Positron, 27 bytes

Positron is a new practical language by @HyperNeutrino.

function{return$1*($1+1)/2}

Try it online!

Answer 57

Perl 6, 11 bytes

{[+] 1..$_}

Try it

{ } creates a lambda block with implicit parameter $_
1 .. $_ creates a Range object
[+] reduces it using the &infix:«+» operator.

(Rakudo actually calls the sum method on the Range object if you haven't lexically modified the &infix:«+» operator, and the sum method knows how to calculate the result without iterating through all of the values)

Answer 58

Element, 32 Bytes

_'1 z;0 t;[z~2:z;t~+t;z~1+z;]t~`

Try it online!

Probably can go shorter, but late now...

Answer 59

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 60

Common Lisp, 26 bytes

(lambda(n)(/(+(* n n)n)2))

Answer 61

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 62

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 63

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 64

Röda, 13 bytes

{seq 1,_|sum}

Try it online!

Answer 65

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 66

Pyt, 1 byte

△

Try it online!

Explanation:

               implicit input
 △             compute the nth triangle number

Answer 67

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

n=>n*++n/2

Try it online!

Answer 68

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 69

BLua, 32 bytes

r=0;for i=1,n r=r+i end;return r

Try it out (Vanilla Lua version, 53 bytes)

I'm not very good at this

Answer 70

CJam, 6 bytes

ri),:+

Try it online!

Explanation

ri    e# Read integer n
)     e# Add 1
,     e# Range from 0 to input argument minus 1
:+    e# Fold addition over array. Implicitly displa

Answer 71

Japt -x, 3 1 bytes

ò

Explanation:

ò     Range [0...Input]
-x    Sum

Try it online!

Answer 72

Add++, 15 9 bytes

D,f,@,Rb+

Try it online!

ಠ_ಠ I forgot about functions. And then I forgot about the range command

How it works

D,f,@,    - Create a monadic function called f (one argument)
      R   - Generate a range from 1 to n
       b+ - Reduce that range by addition (sum)

Answer 73

Braingolf, 1 byte

Q

Try it online!

1-indexed (ie 7 returns 0-6 summed)

Braingolf, 3 bytes

U&+

Try it online!

U - range, &+ - sum.

Answer 74

Clojure, 31 23 16 bytes

8 bytes saved thanks to @cliffroot

#(/(+(* % %)%)2)

Try it online!

Answer 75

AutoHotkey, 22 bytes

A(n){return n*(n+1)/2}

Defines a function that takes parameter n, and returns n*(n+1)/2, which is the nth triangle number, as shown in Leo's Husk answer.

Answer 76

Swift, 13 bytes

Anonymous function:

{$0*($0+1)/2}

You can call it like this:

print({$0*($0+1)/2}(5))

Try it online!

Answer 77

Mathematica, 9 bytes

#(#+1)/2&

or, at 11 bytes,

Tr@Range@#&

Answer 78

Ly, 26 7 bytes

n:1+*2/u

EDIT: Saved a bunch of bytes by using a far better algorithm.

Answer 79

4, 40 bytes

3.70060101002000120300026040230503045054

Try it online!

Answer 80

Tcl, 27 bytes

proc T n {expr $n*($n+1)/2}

Try it online!

Answer 81

Casio Basic, 9 bytes

(n+1)n/2

8 bytes for the code, +1 to add n as parameter.

Answer 82

Factor, 12 bytes

[ iota sum ]

Input is given as an argument to this anonymous function (quotation).

Answer 83

Terrapin Logo, 16 bytes

OP (1+:N)*(:N/2)

Answer 84

GolfScript, 7 6 bytes

~.)2/*

Try it online!

Answer 85

Commentator, 22 bytes

//
;-} {-
 {-  -}<!-}!

Try it online!

Answer 86

><>, 20 15 bytes

:1-:?!v
n?=1l+<

@notatree saved me a couple bytes!

Try it online!

Answer 87

Pip, 7 bytes

a*++a/2

Try it online!

Answer 88

Javascript, 12 bytes

n=>(n*n+n)/2

Answer 89

Carrot, 9 bytes

#^F+1/2*$

Explanation:

#  //Set the string stack to the input
^  //Convert to operations mode
F  //Change to float stack
+1 //Add one to the stack
/2 //Divide the stack by 2
*$ //Multiply the stack by the input
   //Implicitly output the result

Answer 90

MY, 4 bytes

Wow, MY is actually capable of something!

iΣ↵

Try it online!

Explanation (hex/cp):

1A/ - push input as an integer
49/i - pop a; push [1 .. a]
53/Σ - pop a; push sum(a)
27/↵ - pop a; print(a) (with newline)

Answer 91

Befunge, 9 bytes

&:1+*2/.@

Try it online!

Answer 92

Gaia, 2 bytes

+⊢

This is reduce ⊢ by addition +, which implicitly casts numbers to ranges beforehand.

You could also do ┅Σ (range ┅ and sum Σ), which is still 2 bytes.

Answer 93

Excel VBA, 14 Bytes

Anonymous VBE immediate window function that takes input from range [A1] and outputs to the VBE immediate window

?[A1*(A1+1)/2]

Answer 94

Java, 10 bytes

n->n*-~n/2

Admittedly, only slightly different from another Java answer, but uses two's complement trickery.

Answer 95

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 96

Kona - 8 bytes

{+/!x+1}

Explanation:

 +/      Add together
   !      All numbers less than...
    x+1   The input plus 1

Alternative answer - 8 bytes:

{+/x,!x}

 +/      Add together
   x      The input
    ,     Joined to
     !     All the numbers less than...
      x    The input

Answer 97

Emojicode, 50 bytes

➡️➗✖➕1 2

Try it online!

Answer 98

Recursiva, 3 bytes

sBa

Try it online!

Recursiva, 5 bytes

H*a;a

Try it online!

Recursive solution:

Recursiva, 12 bytes

=a0:0!+a#~a$

Try it online!

Answer 99

Cubically, 19 bytes

R3U1F1$:1/1+7*7/0%6

How it works:

R3U1F1              Set the top face to 2
      $             Get the first input as a number
       :1/1+7       Set the notepad to the input + 1
             *7     Multiply the notepad by the input
               /0   Divide the notepad by 2
                 %6 Output the notepad as a number

Answer 100

Pyth, 11 9 bytes

VQ=+ZhN)Z

Explanation:

VQ       For N in range(0, Input)
=+ZhN)   Set Z to Z + 1 + N, suppress output and close function call
Z        Output Z

Try it online!

Answer 101

,,,, 6 bytes

::×+2÷

Try it online!

Explanation:

::     Duplicates the input twice
  ×    Pops off top two values and muliples them
   +   Adds the two values together. (Thus far it's basically n*n+n)
    2÷ Divides by 2

Answer 102

Alumin, 9 bytes

jqdhcpfaf

Try it online!

Explanation

jqdhcpfaf
j          numeric input
 q   p     whlie TOS > 0
  d        duplicate TOS
   hc      subtract 1
      f f  fold over...
       a   ... addition 

Answer 103

SNOBOL4 (CSNOBOL4), 62 40 38 bytes

 N =INPUT
 OUTPUT =N * (N + 1) / 2
END

Try it online!

Answer 104

Pyth - 2 Bytes

sS

Explanation:

sSQ Q added implicitly to resolve arity
s   sum of
 S  integers from one to
  Q input

Answer 105

Kotlin, 13 bytes

{it*(it+1)/2}

Beautified

{
    it * (it + 1) / 2
}

Test

var f: (Int) -> Unit =
{it*(it+1)/2}

fun main(args: Array<String>) {
    println(f(5))
}

TIO

TryItOnline

Answer 106

brainfuck, 14 bytes

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

Try it online!

Takes input as character code, outputs as unary null bytes

Answer 107

Phooey, 10 bytes

&.@+1*/2$i

Try it online!

Explanation

&.@+1*/2$i
&.            write input to the tape
  @           push same input to the stack
   +1         increment tape value
     *        multiply tape value by popped stack value
      /2      divide it by 2
        $i    output as integer

Answer 108

Whitespace, 71 bytes

[S S S N
_Push_0][S N
S _Duplicate_0][T   N
T   T   _Read_STDIN_as_integer][T   T   T   _Retrieve][S N
S _Duplicate][N
S S N
_Create_Label_LOOP][S N
T   _Swap][S S S T  N
_Push_1][T  S S T   _Subtract][S N
S _Duplicate][N
T   S S N
_If_0_Jump_to_Label_PRINT_AND_EXIT][S N
S _Duplicate][S T   S S T   S N
_Copy_2nd][T    S S S _Add][N
S N
N
_Jump_to_Label_LOOP][N
S S S N
_Create_Label_PRINT_AND_EXIT][S N
N
_Discard_top][T N
S T _Print_as_integer]

Letters S (space), T (tab), and N (new-line) added as highlighting only.
[..._some_action] added as explanation only.

Try it online (with raw spaces, tabs and new-lines only).

Example run: input = 3

Command   Explanation                     Stack        Heap   STDIN  STDOUT  STDERR

SSSN      Push 0                          [0]
SNS       Duplicate 0                     [0,0]
TNTT      Read STDIN as integer           [0]          {0:3}  3
TTT       Retrieve input from heap 0      [3]          {0:3}
SNS       Duplicate 3                     [3,3]        {0:3}
NSSN      Create Label_LOOP               [3,3]        {0:3}
 SNT      Swap top two                    [3,3]        {0:3}
 SSSTN    Push 1                          [3,3,1]      {0:3}
 TSST     Subtract (3-1)                  [3,2]        {0:3}
 SNS      Duplicate 2                     [3,2,2]      {0:3}
 NTSSN    If 0: Jump to Label_EXIT        [3,2]        {0:3}
 SNS      Duplicate 2                     [3,2,2]      {0:3}
 STSSTSN  Copy (0-indexed) 2nd (3)        [3,2,2,3]    {0:3}
 TSSS     Add (2+3)                       [3,2,5]      {0:3}
 NSNN     Jump to Label_LOOP              [3,2,5]      {0:3}

 SNT      Swap top two                    [3,5,2]      {0:3}
 SSSTN    Push 1                          [3,5,2,1]    {0:3}
 TSST     Subtract (2-1)                  [3,5,1]      {0:3}
 SNS      Duplicate 1                     [3,5,1,1]    {0:3}
 NTSSN    If 0: Jump to Label_EXIT        [3,5,1]      {0:3}
 SNS      Duplicate 1                     [3,5,1,1]    {0:3}
 STSSTSN  Copy (0-indexed) 2nd (5)        [3,5,1,1,5]  {0:3}
 TSSS     Add (1+5)                       [3,5,1,6]    {0:3}
 NSNN     Jump to Label_LOOP              [3,5,1,6]    {0:3}

 SNT      Swap top two                    [3,5,6,1]    {0:3}
 SSSTN    Push 1                          [3,5,6,1,1]  {0:3}
 TSST     Subtract (1-1)                  [3,5,6,0]    {0:3}
 SNS      Duplicate 0                     [3,5,6,0,0]  {0:3}
 NTSSN    If 0: Jump to Label_EXIT        [3,5,6,0]    {0:3}
NSSSN     Create Label_EXIT               [3,5,6,0]    {0:3}
 SNS      Discard top                     [3,5,6]      {0:3}
 TNST     Print top (6) to STDOUT as int  [3,5]        {0:3}         6
                                                                              error

Program stops with an error: No exit found. (Although I could add three trailing newlines NNN to get rid of that error.)

Answer 109

Ahead, 6 bytes

IEK+O@

Try it online!

Answer 110

@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.

Answer 111

MathGolf, 2 bytes

╒Σ

Try it online!

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

Answer 112

Keg, 2 bytes (SBCS)

Ï⅀

TIO

Keg, 5 bytes (SBCS)

Ï∑+).

Explanation:

Ï#    Range from input to 0. The 0 will not affect the summation.
 ∑+#  Apply all stack: add.
   )# We have to complete the braces if we want to output as an integer.
    .#Output as an integer

TIO

Answer 113

Gol><>, 7 bytes

I:P*2,h

Courtesy of JoKing

Try it online!

8 bytes

I::*+2,h

Try it online!

Answer 114

Go, 29 bytes

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

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

Try it online!

Answer 115

Symbolic Python, 14 bytes

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

Try it online!

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

Answer 116

Shakespeare Programming Language, 154 bytes

(Whitespace added for readability)

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

Try it online!

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

Answer 117

Charcoal, 5 bytes

ＩΣ…·Ｎ

Try it online!

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.

Answer 118

Clojure - 19 bytes

#(apply +(range %))

Boring answer is boring...

Answer 119

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 120

Tcl, 36 bytes

time {incr s [incr n]} $argv;puts $s

Try it online!

Answer 121

Tcl, 29 bytes

puts [expr $argv*($argv+1)/2]

Try it online!

Answer 122

QBIC, 10 bytes

?(:+1)/2*a

This prints (input+1) divided by 2 times input. Looping through all numbers from 1 to n is one byte longer:

[:|p=p+a]?p

Answer 123

Pyke, 2 bytes

Ss

Try it online!

Showing off Pyke's roots in being inspired by Pyth, the function map is exactly the same, only ran as a stack rather than a tree.

S  -  range(1, input+1)
 s - sum(^)

Answer 124

Scala, 12 bytes

Gauss sum (12 bytes):

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

Naive version (14 bytes):

n=>1.to(n).sum

Answer 125

Clojure, 28 bytes

#(apply + (range 1 (inc %)))

Answer 126

Java 8 (39 37 bytes)

n->{System.out.println((n*(n+1))/2);}

saved 2 bytes thanks to @TheLethalCoder. I now realise that I could also omit print statements etc, but there already is a Java answer that I missed, which would end up being pretty much the same thing. Thus I will leave it as this :-)

(Is this the right way of scoring Java? I have no clue if this is valid, I saw it in another answer but this actually omits quite a bit of the code necessary to actually run it..)

try it online

Answer 127

Clojure, 16 bytes

#(/(*(inc %)%)2)

Answer 128

Groovy, 11 bytes

{it*++it/2}

Try it online!

---

The input is not modified by rules of Groovy clsures.

Answer 129

PHP, 35 34 32 25 bytes

<?=$argv[1]*++$argv[1]/2;

Run from the command line, with the input as a parameter.

Answer 130

Fortran 95, 58 bytes

function l(n)
i=1
l=1
do while(i<n)
i=i+1
l=l+i
end do
end

Try it online!

Answer 131

Dyvil, 12 bytes

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

The operator rules force me to use either parentheses or spaces. Uses the Gauss method, and is also a Scala polyglot.

Usage:

let f: int -> int = n=>n*(n+1)/2

print f(5)  // 15
print f(10) // 45

Answer 132

Swift 3, 60 Bytes

var b=Int.init(readLine()!)!
(1..<b).forEach{b+=$0}
print(b)

Try it online!

Answer 133

Math++, 13 bytes

?>n
(n*n+n)/2

Answer 134

Matlab/Octave, 12 bytes

@(n)sum(1:n)

Answer 135

Perl 5, 11 bytes

10 bytes of code + 1 for -p

$_*=$_++/2

Try it online!

Answer 136

Implicit, 2 bytes

¡Þ

Try it online!

    implicit integer input
¡   push 1..n
 Þ  add stack together
    implicit integer output

---

Version without builtins:

Implicit, 13 12 bytes

(:-1)[(]+[)]

The popping of 0 is unnecessary due to updates. Old version:

(:-1);[(]+[)]

Try it online! Explanation:

(:-1);[(]+[)]
(...)           do..while top of stack truthy
 :               duplicate (implicit input if stack empty)
  -1             subtract 1 from top of stack
     ;          pop the last value (will be 0)
      [         pop stack into memory
       (...)    do..while top of stack truthy
        ]       pull memory
         +      add top two stack values together
          [     pop into memory
                implicit output

Ok, so that's a bit confusing. (I don't really know how it works myself). Let's go step-by-step for the input 5.

(:-1);[(]+[)]
(...)
 :-1            stack: 5, 4
 :-1            stack: 5, 4, 3
 :-1            stack: 5, 4, 3, 2
 :-1            stack: 5, 4, 3, 2, 1
 :-1            stack: 5, 4, 3, 2, 1, 0
     ;          stack: 5, 4, 3, 2, 1
      [         stack: 5, 4, 3, 2
       (...)
        ]       stack: 5, 4, 3, 2, 1
         +      stack: 5, 4, 3, 3
          [     stack: 5, 4, 3
        ]       stack: 5, 4, 3, 3
         +      stack: 5, 4, 6
          [     stack: 5, 4
        ]       stack: 5, 4, 6
         +      stack: 5, 10
          [     stack: 5
        ]       stack: 5, 10
         +      stack: 15
          [     stack: empty (exits loop)
             ]  stack: 15
                implicit output

Answer 137

Pushy, 3 bytes

RS#

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How it works

    - Implicit input on the stack e.g. 5;    STACK = [5]
R   - Generate a range from 1 to n;          STACK = [1 2 3 4 5]
 S  - Sum the stack;                         STACK = [15]
  # - Output top of the stack as an integer; DISPLAY 15

Answer 138

Python 3, 35 bytes

print(sum(range(1,int(input())+1)))

Answer 139

C (gcc) 16 bytes

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

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Answer 140

Aceto, 9 bytes

ridIU2:*p

ri reads input and converts to int
d duplicates the value
I increments the top value by one
U reverses the stack
2: divides by 2
* multiplies the top two values
p prints it

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Answer 141

COBOL (GNU), 164 bytes

Requires the -F flag (+2 bytes)
Uses the formula N*(N+1)/2

IDENTIFICATION DIVISION.PROGRAM-ID.S.DATA DIVISION.WORKING-STORAGE SECTION.
01 N PIC 9(9) VALUE ZEROES.PROCEDURE DIVISION.ACCEPT N.COMPUTE N=N*(N+ 1)/2.DISPLAY N.

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Ungolfed:

       IDENTIFICATION DIVISION.
       PROGRAM-ID. S.
       DATA DIVISION.
       WORKING-STORAGE SECTION.
       01 N	PIC 9(5) VALUE ZEROES.
       01 R	PIC 9(9) VALUE ZEROES.

       PROCEDURE DIVISION.
       ACCEPT N
       COMPUTE R=N*(N+ 1)/2
       DISPLAY R.

COBOL (GNU), 362 bytes

Explicitly computing the sum

       IDENTIFICATION DIVISION.
       PROGRAM-ID. S.
       DATA DIVISION.
       WORKING-STORAGE SECTION.
       01 N	PIC 9(5) VALUE ZEROES.
       01 T	PIC 9(5) VALUE ZEROES.
       01 R	PIC 9(9) VALUE ZEROES.

       PROCEDURE DIVISION.
       ACCEPT N.
       PERFORM VARYING T FROM 1 BY 1 UNTIL T > N
       ADD T TO R
       END-PERFORM.
       DISPLAY R.

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Because COBOL uses fixed-size variables, both of these work for values of n up to 44720. Anything larger doesn't fit in the specified 9 decimal digits of the result, although that is easily resolved by changing PIC 9(9) to PIC 9(10).

Answer 142

Windows batch, 57 21 bytes

Inspired by J42161217

@cmd/cset/a%1*(%1+1)/2

Answer 143

Functoid, 29 bytes

Most likely the characters $.@ don't count to the total, but this isn't going to win anyway..

Y(BBxZK0(BBB(C(BBS))CB[+))$.@

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Explanation

Y(                       )     # find fixpoint of the function
  x                            # single argument recursive helper with
   Z                           #   - base case (== 0)
    (K0)                       #   - in base case: constant 0
        (BBB(C(BBS))CB[+)      #   - recursive function ( add (+) argument to recursive call of argument decremented by 1 ([) )
                          $    # apply argument
                           .   # simplify & print as Church numeral
                            @  # exit

The massive combinator BBB(C(BBS))CB is possibly still golfable, I'm still learning how to golf lambda calculus combinators and I find it rather tricky..

Answer 144

Broccoli, 35, 33 bytes

(fn t ($n) (/ (* $n (+ 1 $n)) 2))

Defines a function 't' that can be called like so:

(t 10)

Answer 145

Pyt, 2 bytes

řƩ

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implicit input
ř returns [1, 2, 3, ... , top value]
Ʃ sums top value
implicit output

golfed by mudkip

Answer 146

Clean, 26 bytes

import StdEnv
$n=sum[1..n]

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And also

Clean, 26 bytes

import StdEnv
$n=n*(n+1)/2

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Answer 147

Aheui (esotope), 33 bytes

방빠받반타다따반나망해

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Answer 148

Tidy, 13 bytes

{a:[1,a]|sum}

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Explanation

{a:[1,a]|sum} 
{a:         }    lambda with parameter `a`
   [1,a]         range from 1 to a
        |sum     sum

Same byte count: {n:n*(n+1)/2}

Answer 149

Attache, 8 bytes

Sum@1&`:

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Explanation

This is a composition of two functions: Sum and 1&`:. First, 1&`: is a range from 1 to the input; Sum then sums the elements.

Answer 150

Tir, 4 bytes

{∟+}

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Explanation

{∟+}    a block
 ∟      range from 1 to the input
  +     sum that range

Answer 151

Fynyl, 5 bytes

{rf+}

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Explanation

{rf+}    block
 r       range from 1 to input
  f+     fold addition (sum)

Answer 152

Somme, 6 bytes

n:i*G.

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Explanation

n:i*G.
n         numeric input      [n]
 :        duplicate          [n, n]
  i       increment          [n, n+1]
   *      product            [n(n+1)]
    G     halve              [n(n+1)/2]
     .    output             []

Answer 153

;#+, 19 bytes

*;(~;~*;)-~(~+~;)+p

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Takes input in unary, outputs in decimal.

*;(~;~*;)-~(~+~;)+p
*;                     read 1 byte of input and increment it (check EOF)
   ~;~                 increment the secondary accumulator
  (   *;)              ...while there is still input
         -             set delta value to -1 (subtraction)
                           the state is now (0, N, -)
          ~(~ ~;)      for each character read
             +         subtract N from the secondary accumulator
                           (N decreases with each iteration)
                           the state is now (0, -sum, -)
                 +     subtracts sum from accumulator (0 - (-sum) = 0 + sum = sum)
                  p    print that value

Answer 154

D, 28 bytes

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

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Alternatively, 61 bytes: import std.range;N f(N)(N n){return std.range.iota(n).sum+n;}

Answer 155

DScript, 28 bytes

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

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Alternatively, 34 bytes: N f(N)(N n){return iota(n).sum+n;}

Answer 156

Runic Enchantments, 8 bytes

i:1+*2,@

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Nothing exciting here, reads input multiplies it with itself+1, divides by 2, and outputs.

Kind of feel that I should have anticipated this sort of 1-input-1-output mathematical operation and made a MathFunc (A) operation for it ("its factorial, but addition!"), but I didn't.

Answer 157

W, 2 bytes

+R

Explanation

   % Implicit range from 1 .. input
+R % Reduce the array via addition

Answer 158

Rust, 12 bytes

|n|n*(n+1)/2

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Anonymous function.

Answer 159

Z80Golf, 11 bytes

00000000: cd03 8047 af80 0520 fcff 76              ...G... ..v

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I/O as byte values, takes a byte, outputs a byte. Assembly:

    call $8003
    ld b, a
    xor a
loop:
    add a, b
    dec b
    jr nz, loop
    rst $38
    halt

Z80Golf, 13 bytes

00000000: cd03 805f 193d 20fb 7dff 7cff 76         ..._.= .}.|.v

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I/O as byte values, takes a byte, outputs two bytes little endian. Assembly:

    call $8003
loop:
    ld e, a 
    add hl, de
    dec a
    jr nz, loop
    ld a, l
    rst $38
    ld a, h
    rst $38
    halt

Z80Golf, 54 bytes

00000000: cd03 8038 0ed6 3029 e5d1 2929 195f 1600  ...8..0)..))._..
00000010: 1918 ede5 d11b 197b b220 facd 1f00 7611  .......{. ....v.
00000020: 0000 01f6 ff09 3003 1318 faeb d57d b4c4  ......0......}..
00000030: 1f00 d17b c63a                           ...{.:

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Proper, decimal I/O. Uses a recursive output routine, with a particularly clever 0-byte tailcall:

get_input:
    call $8003
    jr c, got_input
    sub a, '0'
    add hl, hl
    push hl
    pop de
    add hl, hl
    add hl, hl
    add hl, de
    ld e, a
    ld d, 0
    add hl, de
    jr get_input

got_input:
    push hl
    pop de
loop:
    dec de
    add hl, de
    ld a, e
    or a, d
    jr nz, loop

    call output
    halt

; Input:
; HL = number to print
output:
    ld de, 0 ; quotient
    ld bc, -10
div_loop:
    add hl, bc
    jr nc, neg
    inc de
    jr div_loop
neg:
    ex de, hl
    ; now: hl = quotient, de = remainder - 10
    push de
    ld a, l
    or h
    call nz, output
    pop de
    ld a, e
    add '0' + 10
    ; fallthrough to $8000 where the output hook is

Everything assembled with WLA-DX. The -b flag was passed to wlalink, and the following memory map was used:

.ROMBANKMAP
    BANKSTOTAL 1
    BANKSIZE $10000
    BANKS 1
.ENDRO

.MEMORYMAP
    DEFAULTSLOT 0
    SLOTSIZE $10000
    SLOT 0 $0000
.ENDME

Answer 160

Python 3, 15 bytes

sum(range(n))+n

Just a quick check:

>>> n = 6
>>> func = "sum(range(n))+n"
>>> len(func)
15
>>> eval(func)
21