Given sets P = 2n + 1 : n = 1, 2, 3, 4 and Q = 2n − 1 : n = 3, 4, 5, 6, find the intersection P ∩ Q.

The set $P$ is defined as $P = \{2n + 1 : n = 1, 2, 3, 4\}$.
Substitute each value of $n$ into the expression $2n+1$:
For $n=1$: $2(1)+1 = 3$
For $n=2$: $2(2)+1 = 5$
For $n=3$: $2(3)+1 = 7$
For $n=4$: $2(4)+1 = 9$
So, $P = \{3, 5, 7, 9\}$.

The set $Q$ is defined as $Q = \{2n - 1 : n = 3, 4, 5, 6\}$.
Substitute each value of $n$ into the expression $2n-1$:
For $n=3$: $2(3)-1 = 5$
For $n=4$: $2(4)-1 = 7$
For $n=5$: $2(5)-1 = 9$
For $n=6$: $2(6)-1 = 11$
So, $Q = \{5, 7, 9, 11\}$.

The intersection of two sets consists of all elements that are common to both sets.
$P = \{3, 5, 7, 9\}$
$Q = \{5, 7, 9, 11\}$
The elements common to both $P$ and $Q$ are $5, 7,$ and 9.
Therefore, $P \cap Q = \{5, 7, 9\}$.

The final answer is $\boxed{\{5, 7, 9\}}$.