If $X$ ~ Geo $(p)$ and $X_1, X_2$ are independent observations of $X$,
(a) calculate $P(X_1 + X_2 = 3)$,
(b) find $P(X_1 +X_2 =n)$,for $n \ge 2,$
(a) $$ \begin{array}{c|c} X_1 & X_2 \\ \hline 1 & 2\\ 2 & 1 \\ \end{array} $$
$$P(X_1 + X_2 = 3)=2(p \cdot pq) = 2p^2q$$
(b) I don't know how to proceed for (b).
If I write $P(X_1 \ge 2) = pq^1 + pq^2 + \ldots = \frac{p}{1-q}$ Am I on the right track?  Following an answer posted by a member in this community ,
\begin{align} P(X_1+X_2=n) &= \sum_{i=1}^{n-1} P(X_1=i)P(X_2=n-i) \\ &= \sum_{i=1}^{n-1} q^{i-1}pq^{n-i-1}p \\ \end{align}
I proceeded as follows:
$\frac{p^2}{q^2} \sum ^n _{i=2} q^n$
This simplified to $p$ when the answer is $pq^{n-4}$
$\frac{p^2}{q^2} \sum ^n _{i=0} q^n - (1+q)$
$ \frac{p^2}{q^2} [\frac{1}{1-q} -(1+q)]$
\begin{align} P(X_1+X_2=n) &= \sum_{i=1}^{n-1} P(X_1=i)P(X_2=n-i) \\ &= \sum_{i=1}^{n-1} q^{i-1}pq^{n-i-1}p \\ \end{align}
Can you collect the terms and simplify the equation?
Edit:
\begin{align} P(X_1+X_2=n) &= \sum_{i=1}^{n-1} P(X_1=i)P(X_2=n-i) \\ &= \sum_{i=1}^{n-1} q^{i-1}pq^{n-i-1}p \\ &=\sum_{i=1}^{n-1} p^2q^{n-2} \\ &=(n-1)p^2q^{n-2} \end{align}