It is faster and less error prone to use the summation formula rather than doing the brute force summation.

use strict;
use warnings;
my $number = $ARGV[0];
my $sum_of_squares = sub{
    my $n = shift;
    return $n * ($n + 1) * (2*$n +1) / 6;
};

print $sum_of_squares->($number);
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This formula is available in many places. I first found it at https://brilliant.org/wiki/sum-of-n-n2-or-n3/.