How silly. To calculate the numbers, you would use n(n+1)/2, no loop required. Maybe the loop is to ask for more input. :P

Anyway, it sounds like the requirements for this work are: get a command line argument, create a loop, and although not mentioned, print it out. It should be easy enough to write this program. Though, if you are having any difficulty, show us your work, and perhaps we can help. :)

$n = 13; 
$sum = 0; 
$sum += $n-- while $n; 
print $sum;;
91
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The following calculates the sum efficiently (sum(1..N) = N(N+1)/2), yet finds a legitimately use for foreach:

foreach my $n (@ARGV) { print($n*($n+1)/2, "\n"); }
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Agent 006 undercover ;)

Cheers Rolf
_{(addicted to the Perl Programming Language and ☆☆☆☆ :)}

PS: Je suis Charlie!

Perl 6 mojo?

We told you guys; it’s fun! Though I’m not positive the answer shouldn’t have been 91 instead… to v through.

~>perl6
> $n = 13
13
> say [+] ^$n
78
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"...mojo?"

Or Voodoo? Perhaps a servant of the spirits?

Best regards, Karl

«The Crux of the Biscuit is the Apostrophe»

Just do it in a slightly more complicated manner than is strictly necessary...

#!/usr/bin/perl

use strict;
use warnings;
use feature qw/say/;

while(my $n = shift @ARGV) {
    say
        grep    {  $n == $_  }
        map     {  $n += $_  }
        map     {  $n--      }
        map     { ($n) x $_  }
                   $n

;}
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use strict;
use warnings;

my $sum = 0;

while(1){
    print "Please input a number: ";
    chomp(my $number = <STDIN>);

    add $number to $sum;
    print "Your sum is now: $sum";
}
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use strict;
use warnings;

print "Please input your string of numbers seperated by "+";

my @array = split(/[+]/, <STDIN>);
my $sum = 0;

foreach(@array){
add each number to the $sum
}
print "This is your sum: $sum";
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use strict; use warnings; use 5.016;

my $n = $ARGV[0] or die "Give a damn! Give a number.";

require List::Util;
say List::Util::sum( 1 .. $n );
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Now with Extra Fracking for-Loop!

...
say List::Util::sum( 1 .. $n ) for 0;
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Here is a simulation approach:

use strict;
use warnings;

my $n = shift;
my $sum = 0;
my $count = 0;
while( $sum != ($n+1)*$n/2 ) {
    $sum = 1 + int( rand( $n**2 ) );
    $count++;
}
print "Sum of 1 to $n: $sum (found after $count iterations)\n";
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"We don't do homework for you..." (Anonymous Monk)

This ain't true, as you can see - but it may be the provided solutions don't meet your specs sometimes ;-)

So i feel free to put in my 2 ˘:

Edit: In German this sounds much better: Ich bin so frei, meinen Senf dazu zu geben.

#!/usr/bin/env perl

use strict;
use warnings;
use List::Util qw(reduce);
use feature qw (say);

my $n = shift || die qq(Usage $0 [number]);

say reduce { $a + $b } 1 .. $n;

__END__
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Please see also There's more than one way to do it as well as List::Util.

Best regards, Karl

WinZ>perl -e " print eval join '+',(1..$ARGV[0]);exit while 1" 13
91
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This response is not intended to be taken literally. :p

The sum of ALL numbers being negative 1/12, would the sum of all numbers up to N not be some type of a function? In the manner of

Oh yes, I see now n(n+1)/2, overthinking it again.

Although that might be N * -1/12 (but is probably neither). For another undefined definition, maybe just bitmasking the sum of all numbers to length N. This could work well if soan is truly irrational.

sub donotrun_this_is_an_example{
  
 use constant Soan = (-1/12);
 my $bitmask = ~ -256**($n/8);

return Soan & $bitmask;
}
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similar idea would be to return the of multiplying Soan by 10 to power of N: int( Soan*10**$n )

Edit: Soan is negative 1/12