Iterate over arguments in a bash script and make use of their numbers

Question

If I want to iterate over all arguments it is as easy as for i in "$@"; do .... However, let's say I want to start with the second argument and also make use of the arguments' positions for some basic calculation.

As an example I want to shorten these commands into one loop:

grep -v 'foobar' "$2" | grep -f $file > output1.txt
grep -v 'foobar' "$3" | grep -f $file > output2.txt
grep -v 'foobar' "$4" | grep -f $file > output3.txt
grep -v 'foobar' "$5" | grep -f $file > output4.txt

I tried many variations like for i in {2..5}; do grep -v 'foobar' "$$i" | grep -f $file > output$(($i-1)).txt; done; however, it seems bash expansion doesn't work like this.

EDIT:

Seems I made a mistake not emphasizing that I need to make use of the argument's position/number (i.e., 2 from $2). It's important because the output files get used separately later in the script. All of the provided answers so far seem correct but I don't know how to use them to make use of the argument's "number".

Answer 1

Couple correct answers already, another way could be:

for (( i=2; i <= "$#"; i++ )); do
    echo "arg position: ${i}"
    echo "arg value: ${!i}"
done

Answer 2

If you do not want to shift off the first unneeded arguments you can use the indirection expansion:

for i in {2..5}; do
  echo "${!i}"
done