= Solution
A coordinate $Q_s$ of an $A_{1g}$ <normal mode> is unchanged by every <molecular symmetry operation>. Every harmonic-oscillator excitation $H_n(Q_s)e^{-Q_s^2/2}$ therefore also has symmetry $A_{1g}$, including odd values of $n$. Consequently
$$
\boxed{B_{2u}\otimes E_{2g}\otimes A_{1g}=E_{1u}}
$$
for any number of quanta in that totally symmetric coordinate. \b[The same electric-dipole permission survives every $A_{1g}$ excitation.]
This statement concerns symmetry. Actual progression intensities depend on vibrational <inner products>, equilibrium displacements and the coordinate dependence of the <electric dipole transition moment>; a permitted line need not have measurable intensity for every vibrational quantum number.
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