How do I find $P(X+Y = k)$ for a geometric distribution? If $X$ and $Y$ are independent identically distributed random variables where 

$P(X=k) = P(Y=k) = pq^{k-1}$ where $q = 1-p$.

How do you find $P(X+Y=k)$? Is it acceptable to say that
$$P(X+Y=k) = P(\{X=m\} , \{Y=k-m\}) = P(X=m)P(Y=n-m)$$ since$ X, Y$ are independent? I'm not sure if this is legitimate and I'm also not sure whether it's possible to do this without throwing in the extra '$m$' variable. 

Any advice would be much appreciated. 

Thanks 

The wider question that I'm trying to solve is $P(X=k \mid X+Y=n+1)$ and my method so far requires $P(X+Y = n+1)$ - is there another method?
By its moment generating function, you can prove that the sum of geometric distribution rv is a negative binomial distribution.

your way is legit. but for a k of large value, you have to calculate a lot.  

please refer to this link:Sum of two independent geometric random variables
By its moment generating function, you can prove that the sum of geometric distribution rv is a negative binomial distribution.

your way is legit. but for a k of large value, you have to calculate a lot.  

please refer to this link:Sum of two independent geometric random variables