According to the Access lecture, which was conducted by the chief examiner, you need to account for multiplets. It was about the only useful piece of information I gleaned that entire night.
Multiplets or multiplicity? Multiplets are just any peaks that aren't singlets afaik...
Multiplets are peaks that cannot be distinguished as a simple n-tet or (tet)
2, and so forth.
The
\times (n_2+1))
rule applies if you have different splitting frequencies. However, you need to consider if the difference is significant enough to break the typical (N+1) coupling. If the two neighbouring environments are similar, then peaks in
\times (n_2+1))
splitting will overlap each other, and give you the typical (N+1) coupling. Even if difference in splitting frequencies is significant, if the instrument is not sensitive enough, you won't get your
\times (n_2+1))
splitting (instrumental broadening is always a tricky and expensive problem to fix. UoM's NMR machine costs ~$1M). Typically, you only see this kind of splitting in aromatic compounds, as the delocalised ring allows 4-bond coupling (i.e. second neighbours), which has a very different J-coupling value compared to 3-bond coupling. Of course, none of this is taught in VCE (I learnt this in 2nd year), so there is no way to ask you to do this
Also, if the difference is significant enough, and you have for example (1+1)*(2+1)=6 peaks, that is NOT a sextet. That is a doublet of triplets or triplet of doublets, depending on the strength of J-coupling. To name something n-tet, it must have the correct binomial height distribution (i.e. Pascal's triangle).
\times (n_2+1))
splittings look like this:

Anyways, the point is you only need (n+1) splitting. A case can always be made that
\times (n_2+1))
splitting is insignificant enough to be ignored.