Vtyjkyuk

$m$ is the size of the hash table and $n$ is the number of entries in the hash table.

We assume the hash function uniformly and independently distributes the keys among the values $0$ to $m-1$.

In Robert Sedgewick's book Algorithms 4th ed. He writes

If a cluster is of length $t$, then the expression $(t+(t-1)+ldots
+2+1)/m=t(t+1)/(2m)$ counts the contribution of that cluster to the grand total.
The sum of the cluster lengths is $n$, so, adding this cost for all
entries in the table, we find that the total average cost for a search
miss is $1+n/(2m)$ plus the sum of the squares of the lengths of the
clusters, divided by $2m$. Thus, given a table, we can quickly compute
the average cost of a search miss in that table.

My question is why do we add $n/(2m)$ to the expected number of probes. I understand the $1$ is added because every entry requires at least one probe, and adding the square of the length of a cluster (divided by $(2m))$ accounts for a search miss beginning within that cluster, but I don't see how the $n/(2m)$ is relevant.