I was trying to prove the expected value of $X$ when $X\sim NB(n,p)$, then I came across this proof: for geometric distribution $X_1, X_2,\ldots, X_n$ with the same parameter $p$,
(At least I think) I get the idea; since NBD is focusing on the number of trials before we get $n$ successes, it would be reasonable to think of it as a combination of $n$ independent GDs.
But we need at least $n$ trials for $n$ successes, whereas only 1 trial is enough for each $X_i$. That is, $X$ prevents $n$th success from occuring before $n$th trial, but it is not the case for $X_i$s (since they are all independent).
Why is it okay to express $X$ as $\sum X_i$?