 Solution

Reflection $x\mapsto a-x$ preserves the walls and kinetic operator but sends $\lambda (x-a/2)$ to $-\lambda (x-a/2)$. The Hamiltonians at $\lambda$ and $-\lambda$ are unitarily equivalent, so each analytic nondegenerate energy branch obeys $E_n(\lambda )=E_n(-\lambda )$. Its Taylor expansion contains only even powers: all odd-order energy coefficients vanish, not merely the first-order expectation.