Can checking additional congruencies be done for the strong version of Lucas test? Relating to how the Lucas sequence can be used to check if a number is prime or not, I've red various "additional congruence conditions" that must hold true if the number is prime. The Wikipedia article states

there are additional congruence conditions we can check that have almost no additional computational cost 

Take for example the additional check $V_{n+1} \equiv 2Q (\mod{n})$
For the strong version of the Lucas test, this certainly is not "at no additional cost" (see here). It does not calculate up to the $n+1$ term. Are these additional congruency checks only for the normal (aka weak) version of the Lucas test which does go up to $n+1$? Does it make sense to check additional congruencies when using the strong form of the test or would it not make a difference?
Another example from the article is 

Recall that $Q^{n + 1}$ is computed during the calculation of $U_{n + 1}$. 

But $U_{n + 1}$ isn't computed for the strong version of the test. 
 A: Well, the additional checks always carry some additional cost, whichever test you're doing - the question is how much.
The reason that they make sense to do here is that the additional cost is a lower-order term. Computing $V_{n+1}$ takes $O(\log n)$ arithmetic operations. Comparing it to $2Q$ modulo $n$ takes only $O(1)$ arithmetic operations. For large $n$, computing $V_{n+1}$, which we had to do anyway, takes the majority of the time.
It's true that if the strong Lucas test is used, then $V_{n+1}$ would not have been computed: we might have stopped at $V_{(n+1)/2}$, or $V_{(n+1)/4}$, or $V_{(n+1)/8}$, or something even smaller. However, this isn't a significant difference, because for most values of $n$ we could consider, the highest power of $2$ dividing $n+1$ isn't all that high. For instance, more than $99.9\%$ of the time, $n+1$ is not divisible by $2^{10} = 1024$, and so the strong Lucas test won't be more than $10$ steps away from getting to $V_{n+1}$.
If you're using the strong Lucas test, you can still do the extra checks by simply finishing the job and getting all the way to $V_{n+1}$ even though you didn't need it. True, this means that there is a greater additional cost to the extra checks - but in expectation, it's still an $O(1)$ cost compared to the $O(\log n)$ cost of doing the test to begin with. 
Very rare inputs $n$ such as Mersenne numbers might be an exception to this rule, where the strong Lucas test might be very far away from computing $V_{n+1}$, and so the additional cost of an extra check is much worse.
