$X,Y$ distributes geometrically, calculate $P(X+Y=n)$? I have this question (Answer provided).



I'm asking only about $P(X+Y=n)$

My calculation is:
$$\sum_{k=0}^n P(X = k, Y = n-k) $$ 
And when calculated like this, the final answer is: $(n+1) p^2 (1-p)^n$

I've searched online and found that in a lot of places $(n-1)$ will recieved if the sigma is: $$\sum_{k=1}^{n-1}$$ So it might be the key to the answer.

BUT, I don't have any constraints in the question so why should it be like this? 
Thanks
There are two different definitions of Geometric distribution (check wikipedia or anything else). https://en.m.wikipedia.org/wiki/Geometric_distribution

You use the one which starts from zero, authors of this problem use the one that starts from 1.
The geometric distribution has probability mass zero on $0$.  Consequently, 
$$ P(X = 0, Y = n) = P(X = n, Y = 0) = 0  \text{.}  $$

Therefore, your 
$$ \sum_{k=0}^n P(X = k, Y = n-k) = 0 + \left( \sum_{k=1}^{n-1} P(X = k, Y = n-k) \right)  + 0  \text{.}  $$