Suppose that X and Y are independent, identically distributed, geometric random variables with parameter p. Now I have to show that: 
P(X = i|X + Y = n) = $\frac{1}{n-1}$ ,for i = 1, 2, ..., n − 1.
I have managed to go this far:
P(X = i| X + Y = n) = $\frac{P(X = i | X + Y = n)}{P(X + Y = n)}$ = $\frac{P(X = i)P(Y = n − i)}{P(X + Y = n)}$  
P(X = i) = $p(1 − p)^{i-1}$
and
P(Y = n − i) = $p(1 − p)^{n−i−1}$
It follows that
P(X = i)P(Y = n − i) = $p^2(1 − p)^{n−2}$
and then from here on I have no idea what to do to find what the question has asked for.
 A: You are already almost done:  Note that the joint probability $$\Pr[(X = i) \cap (Y = n-i)] = p^2 (1-p)^{n-2}, \quad i = 1, 2, \ldots, n-1,$$ is constant with respect to $i$.  So $$\Pr[X + Y = n] = \sum_{i=1}^{n-1} \Pr[(X = i) \cap (Y = n-i)] = (n-1) p^2 (1-p)^{n-2},$$ because all of the terms in the sum are the same, and there are $n - 1$ such terms.  The result immediately follows.
A: Your approach is good; all that's left is to figure out the distribution of $X + Y$.
Recall that Geometric random variables describe the probability of having the first success of a Bernoulli experiment with parameter $p$ after exactly $n$ trials. Hence, $X + Y$ describes the probability of having the $second$ success of such an experiment after $n$ tries. What does this mean?
Well, we must have $(n-2)$ fails and $1$ success in the first $(n-1)$ trials, and a success on the last one. Keep in mind that there are $(n-1)$ possible trials on which the first success can occur. Hence,
$$P(X+Y = n) = (n-1)p(1-p)^{n-2}p = (n-1)p^2(1-p)^{n-2}.$$
Combining this with your previous result, we obtain
$$P(X=i | X + Y = n) = \frac{p^2(1-p)^{n-2}}{(n-1)p^2(1-p)^{n-2}} = \frac{1}{n-1}.$$
