Sum of independent discrete random variables Let $X_n$ be a sequence of independent random variables with distribution:
$$
P(X_i = -1)=q, \hspace{1cm}
P(X_i = 0) = 1-2q = r, \hspace{1cm}
P(X_i = 1) = q
$$
for $i=1,..,n$ and $q \in (0,1/2)$.
I am investigating the distribution of $S_n = X_1 + X_2 + ... + X_n$.
What I have done for far:
let $p_n:= P(S_n = 0)$, then:
\begin{equation}
p_n = r^n + n(n-1)q^2  p_{n-2}.
\end{equation}
Since all of the random variables $X_1, ..., X_n$ will be 0 or, the second possible option is that two of them will equal $1$ and $-1$ and the rest n-2 random variables will sum up to 0.
Now we calculate $p_{n-2}$ the same way:
\begin{equation}
p_{n-2} = r^{n-2} + (n-2)(n-3)q^2 p_{n-4}.
\end{equation}
(1) and (2) combined together give us:
$$
p_n = r^n + n(n-1)q^2  p_{n-2} =r^n + n(n-1)q^2  [ r^{n-2} + (n-2)(n-3)q^2 p_{n-4}]  = 
$$
$$
r^n + \frac{n!}{(n-2)!}q^2 r^{n-2} +  \frac{n!}{(n-4)!}q^4 p_{n-4} = .... = r^n + \frac{n!}{(n-2)!}q^2 r^{n-2} + ... + \frac{n!}{(n-2k)!}q^{2k} r^{n-2k} + ... .
$$
Analysing that we can deduce that:
$$p_n = P(S_n=0) = 
 \left\{ \begin{array}{ll}
\sum \limits_{k=0}^{n/2} \frac{n!}{(n-2k)!} q^{2k}r^{n-2k} & \textrm{when n is even},\\
\sum \limits_{k=0}^{\frac{n-1}{2}} \frac{n!}{(n-2k)!} q^{2k}r^{n-2k} & \textrm{when n is not even}.\\
\end{array} \right.
$$
What seems to be accurate. I've checked for $n=1,2,3$ - so there is a chance that this calculation is correct.
Then I tried to calculate $P(S_n=i)$  as 
$$
P(S_n=i) = \binom{n}{i} q^i P(S_{n-i} = 0)
$$
for $i>0$,
as we need to have at least i ones (and we have to pick which $X_i$ are equal 1) and the rest (n-i) together must sum up to 0,
and
$$
P(S_n=i) = \binom{n}{-i} q^{-i} P(S_{n+i} = 0),
$$
for $i<0$, but unfortunately this is incorrect. 
I've checked in R that for $q=1/8$ and $n=3$:
$$
\sum \limits_{i=-n}^{i=n} P(S_n = i) = 1.011719 \neq 1
$$
Please help me to find a mistake in my solution or help to calculate $P(S_n =i)$ in a  different way.
 A: On base of non-negative integers $u,v,w$ with $u+v+w=n$ you could start with:
$$\Pr(U_n=u\wedge W_n=w)=\frac{n!}{u!v!w!}q^{u+w}r^v=\frac{n!}{u!(n-u-v)!w!}q^{u+w}r^{n-u-w}$$
where $U_n$ stands for the number of times that $-1$ and $W_n$ for the number of times that $1$ shows up.
Then: $$\Pr(S_n=i)=\sum_{w-u=i}\Pr(U_n=u\wedge W_n=w)$$ 

addendum:
If $B_n^{(i)}$ are iid for $i=1,2$ with Bernouilli($p$) distribution where $p$ is a root of the equality: $$p(1-p)=q$$
then $X_n$ and $B_n^{(1)}-B_n^{(2)}$ have the same distribution. 
Defining $T_n^{(i)}=B_1^{(i)}+\cdots+B_n^{(i)}$ for $i=1,2$ we find that $S_n$ and $T_n^{(1)}-T_n^{(2)}$ have equal distribution, while $T_n^{(i)}$ are iid and have binomial distribution with parameters $n$ and $p$.
This observation can help by for instance finding things like the variance or characteristic function of $S_n$.
A: I advise you to use generating functions. Let $Y_i=X_i+1$ (in order to work with positive valued random variables). In this way, because of independance, $S_n+n$ has generating function:
$$(q+rs+qs^2)^n=(s+q(1-s)^2)^n$$
It suffices to expand this expression in order to get the general term (don't forget to subtract $n$ at the end). In this way, one will find "trinomial coefficients" with some similarity with coefficients in the answer by @drhab.
