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.

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• Closely related – FryAmTheEggman Jul 18 '17 at 20:36
• @FryAmTheEggman Sorry - had a bit of a brain fart there. I see what you mean. – GarethPW Jul 18 '17 at 20:45
• @Aaron you got ninja'd by Husk, which was just posted with a 1 byte solution – Skidsdev Jul 18 '17 at 21:35
• I suggest a stack snippet. – programmer5000 Jul 19 '17 at 11:42
• – Jerry Jeremiah Jul 27 '17 at 12:20

4, 40 bytes

3.70060101002000120300026040230503045054


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Braingolf, 1 byte

Q


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

Braingolf, 3 bytes

U&+


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U - range, &+ - sum.

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


MY, 4 bytes

Wow, MY is actually capable of something!

𝕫iΣ↵


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Explanation (hex/cp):

1A/𝕫 - 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)


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


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• Beat you by 7 bytes – Tornado547 Dec 11 '17 at 16:38
• @Tornado547 I know, but that answers been taken. This is just a unique way of doing it. 2 bytes: sS. – Stan Strum Dec 11 '17 at 16:39

,,,, 6 bytes

::×+2÷


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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


Add++, 15 9 bytes

D,f,@,Rb+


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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)


Recursiva, 3 bytes

sBa


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Recursiva, 5 bytes

H*a;a


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Recursiva, 12 bytes

=a0:0!+a#~a$ Try it online! Pyth - 2 Bytes sS  Explanation: sSQ Q added implicitly to resolve arity s sum of S integers from one to Q input  SNOBOL4 (CSNOBOL4), 6240 38 bytes  N =INPUT OUTPUT =N * (N + 1) / 2 END  Try it online! 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  • @Riker shoot lol done – Conor O'Brien Dec 17 '17 at 21:06 brainfuck, 14 bytes ,[[>+<<.>-]>-]  Try it online! Takes input as character code, outputs as unary null bytes Japt-x, 3 1 bytes ò  Explanation: ò Range [0...Input] -x Sum  Try it online! Phooey, 10 bytes &.@+1*/2$i


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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


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 (╒)

Keg, 3 bytes (SBCS)

Ï⅀.


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

• That's available on the github interpreter and theoretically, it should work. – Jono 2906 Oct 13 at 1:13

Tcl, 36 bytes

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


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Tcl, 29 bytes

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


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• If you've got two Tcl answers, you should probably put them in one answer. – numbermaniac Jul 19 '17 at 0:24
• @numbermaniac I have three. – sergiol Jul 19 '17 at 0:29
• Ah. It's probably better to put them all in one answer, then. – numbermaniac Jul 19 '17 at 1:08

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


Pyke, 2 bytes

Ss


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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(^)


Scala, 12 bytes

Gauss sum (12 bytes):

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


Naive version (14 bytes):

n=>1.to(n).sum


Clojure, 28 bytes

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


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..)

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• (n) can be just n. You can return from the lambda, I believe, instead of printing. And yes it's fine to compile to a lambda. – TheLethalCoder Jul 19 '17 at 12:34
• Side note: Use four spaces to format the code not back ticks. – TheLethalCoder Jul 19 '17 at 12:34
• @TheLethalCoder thanks! I just found another Java answer to this thread which indeed does it a great deal shorter and does not print ^^ – Dylan Meeus Jul 19 '17 at 12:34

Clojure, 16 bytes

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


Groovy, 11 bytes

{it*++it/2}


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The input is not modified by rules of Groovy clsures.

PHP, 353432 25 bytes

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


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

• Save 6 bytes with \$argn and -F – Titus Jul 26 '17 at 11:34

Fortran 95, 58 bytes

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


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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