<s>[INST] Find the remainder of $\frac{3^{11}-1}{2}$ divided by $9$ I have the following question:

Find the remainder of $$\frac{3^{11}-1}{2}$$ divided by $9$

I tried to reformat the question:
$$\frac{3^{11} -1 } {2} \times \frac{1}{9} = \frac{3^{11}-1}{18}$$
Since $3^2 = 9$
$$\frac{3^2(3^9) -1}{3^2 \times2}$$
I don't know where to go next. Anyway, this is one of my many attempts to solve this question, and most of them ends with a complicated solution. I don't want to use modular arithmetic for this question. A hint or anything will help me. 
 [/INST] $\frac {3^{11} - 1}{2} = \frac {3^{11} - 1}{3-1}=$
$3^{10} + 3^9 + .... + 3^2 + 3 + 1=$
$9(3^8 + ..... + 1) + 4$ 
so the remainder is $4$.
..or...
$\frac {3^{11}-1}2 = 3^{11}*\frac 12 - \frac 12 \equiv k \mod 9$
$ 3^{11}-1 \equiv 2k \mod 9$
$2k + 1 \equiv 0 \equiv 9 \mod 9$
$2k \equiv 8 \mod 9$
$k \equiv 4 \mod 9$.
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