Separate Chaining hashing: time complexity of successful search
The expected length of a chain when the $i$th element is added is $\frac{i-1}m$. So if we mentally re-order the elements in the hash map to be in order of insertion, we can calculate the expected length of a chain $E[l]$ when performing a search with $n$ elements in the map:

\begin{align}
E[l]&amp;=\frac{1}{n}\sum_{i=1}^n \frac{i-1}m = \frac{1}{nm}\left(-n + \sum_{i=1}^n i\right) = \frac{1}{nm}\left(-n + \frac{1}{2}n(n+1)\right)\\
E[l] &amp;= \frac{1}{m}\left(-1 + \frac{1}{2}n+\frac{1}{2}\right) = \frac{n}{2m} - \frac{1}{2m}
\end{align}

Now you can substitute $\alpha = \frac n m$ and find $E[l] = \frac \alpha 2  - \frac{\alpha}{2n}$.