Is there a shortcut? I already have E[X] where X=Y+1, thus I can separate it into E[Y]+E[1], and for Y, it's number of tries is n, and the probability p.
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$\begingroup$As long as $X$ is nonzero, you should be able to apply the Law of the Unconscious Statistician to compute
$E[1/X] = \sum_j \frac{1}{x_j} p(x_j)$
or in the continuous case
$E[1/X] = \int \frac{1}{x} p(x) dx$.
Edit: it seems OP is looking for the case when Y~bino(n,p). So from the definition and the law we have $E[1/X] = E[1/(Y+1)] = \sum_{j=0}^n \frac{1}{j+1} p(n,j)$ where $p(n,j) = \binom{n}{j}p^j(1-p)^{n-j}$. I'll leave the fun part of computing the sum up to OP!
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